<\/span><\/h2>\nQuestion 1.
<\/strong>Draw the graph of each of the following linear equations in two variables.<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.1.png\")
Solution:
<\/strong>(i)<\/strong>x + y = 4
x = 4 \u2013 y
If y = 0, then x = 4
If y = 4, then x = 0
Now plot the points (4, 0) and (0, 4) on the graph and join them to get the graph of the given equation
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.2.png\")
(ii)<\/strong>x \u2013 y = 2
x = 2 +y
If y = 0, then x = 2 and if y = 1,
Then x = 2 + 1 = 3
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.3.png\")
Now plot the points (2, 0) and (3, 1) on the graph and join them to get the graph of the equation.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.4.png\")
(iii)\u00a0<\/strong>-x+y = 6 \u21d2\u00a0 y = 6+x
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.5.png\")
If x = 0, then y = 6 + 0 = 6
If x = -1, then y = 6 \u2013 1 = 5
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.6.png\")
Now plot the points (0, 6) and (-1, 5) on the graph and join them to get a graph of the line.
(iv)<\/strong>\u00a0y = 2x
If x = 0, then y =\u00a0 2 x 0 = 0
If x = 1, then y = 2 x 1 = 2
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.7.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.8.png\")
Now plot the points (0, 0) and (1, 2) on the graph and join them to get the graph of the line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.9.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.10.png\")
Now plot the points (5, 0) and (0, 3) on the graph and join them to get the graph of the line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.11.png\")
Now plot the points (4, 0) and (2, -3) on the graph and join them to get the graph of the line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.12.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.13.png\")
Now plot the points (-1, 2) and (2, 3) on the graph and join them to get the graph of the line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.14.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q1.15.png\")
Now plot the points (1, 0) and (-1, 1) on the graph and join them to get the graph of the line.<\/p>\nQuestion 2.
<\/strong>Give the equations of two lines passing through (3, 12). How many more such lines are there, and why?
<\/strong>Solution:
<\/strong>\u2235\u00a0 Points (3, 12) lie on the lines passing through the points
\u2234 Solutions is x = 3,y- 12
\u2234\u00a0 The possible equation can be
x + y = 15
-x+y = 9
4x-y = 0
3x \u2013 y + 3 = 0<\/p>\n\u00a0<\/div>\n
Question 3.
<\/strong>A three-wheeler scooter charges \u20b915 for the first kilometer and \u20b98 each for every subsequent kilometer. For a distance of x km, an amount of \u20b9y is paid. Write the linear equation representing the above information.
<\/strong>Solution:
<\/strong>Charges for the first kilometer = \u20b915
Charges for next 1 km = \u20b98
Distance = x km
and total amount = \u20b9y
\u2234 The linear equation will be,
15 + (x- 1) x 8 =y
\u21d2\u00a0 15 + 8x \u2013 8 = y
\u21d2\u00a0 \u00a07 + 8x = y
\u2234\u00a0 y = 8x + 7<\/p>\nQuestion 4.
<\/strong>Plot the points (3, 5) and (-1, 3) on graph paper and verify that the straight line passing through these points also passes through the point (1, 4).
<\/strong>Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q4.1.png\")
Points (3, 5) and (-1, 3) have been plotted on the graph and joined to get a line. We see that die point (1,4) also lies out.<\/p>\nQuestion 5.
<\/strong>From the choices given below, choose the equation whose graph is given in the figure.<\/strong>
(i) y = x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(ii) x + y = 0<\/strong>
(iii) y = 2x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(iv) 2 + 3y = 7x
<\/strong>Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q5.1.png\")
From the graph, we see that Points (-1, 1) and (1, -1) are on the graph of the line these will satisfy the equation of the line
\u2234\u00a0 -x = y \u21d2 x+ y = 0
i.e., the required equation
\u2235 x + y = 0 is the graph of the equation.<\/p>\nQuestion 6.
<\/strong>From the choices given below, choose the equation whose graph is given in the figure.<\/strong>
(i) y = x + 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(ii) y = x \u2013 2<\/strong>
(ii) y = -x + 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(iv) x + 2y = 6
<\/strong>Solution:
<\/strong>From the graph
Points (-1,3) and (2, 0) lie on the graph of the line
Now these points, by observation, satisfy the equation y= -x+2
\u2234 The required equation is y = -x + 2 whose graph is given.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q6.1.png\")
<\/p>\nQuestion 7.
<\/strong>If the point (2, -2) lies on the graph of the linear equation 5x + ky =4, find the value of k.
<\/strong>Solution:
<\/strong>\u2235\u00a0 Point (2, -2) lies on the graph of the linear equation 5x + ky = 4
\u2234\u00a0 It will satisfy it
\u2234 Now substituting the values of x and y 5 x 2 + k (-2) = 4
\u21d2 10 \u2013 2k = 4 \u21d2\u00a0 -2k = 4 \u2013 10 = -6 -6
\u21d2 k=\u00a0\u2212<\/span>6<\/span><\/span>\u2212<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0=3
Hence k = 3<\/p>\nQuestion 8.
<\/strong>Draw the graph of the equation 2x + 3p = 12. From the graph find the coordinates of the point.<\/strong>
(i) whose y -coordinates are 3<\/strong>
(ii) whose x-coordinates are -3<\/strong>
Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q8.1.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q8.2.png\")
Plot the points (6, 0) and (0, 4) on the graph and join them to get the graph of the line.
(i)<\/strong> If y = 3, then draw a perpendicular from y = 3 to the line, which gets meets it at P then the x-coordinate of p will be
\u2234 coordinates of P are (\u00a03<\/span>2<\/span><\/span><\/span><\/span><\/span>,3)
<\/span>(ii)<\/strong>\u00a0If x = -3, draw perpendicular from x = -3 to the line, which meets it Q.
The y coordinates of Q will be y = 6
\u2234 co-ordinates of Q are (-3, 6)<\/p>\nQuestion 9.
<\/strong>Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the\u00a0coordinates axes:<\/strong>
(i) 6x \u2013 3y = 12\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(ii) -x + 4y = 8<\/strong>
(iii) 2x + y = 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(iv) 3x + 2y + 6 = 0
<\/strong>Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q9.1.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q9.2.png\")
Now plot the points of each equation and join then we get four lines as shown on the graphs.
Equation (i) cuts the axes at (2, 0) and (0, -4)
Equation (ii) cuts the axes at (-8, 0) and (0, 2)
Equation (iii) cuts the axes at (3, 0), (0, 6) and
Equation (iv), cuts the axes at (-2, 0) and (0,-3)
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q9.3.png\")
<\/p>\nQuestion 10.
<\/strong>A lending library has fixed charges for the first three days and an additional charge for each day thereafter. Aarushi paid \u20b927 for a book kept for seven days. If fixed charges are \u20b9x and per day charges are \u20b9y. Write the linear equation representing the above information.<\/strong>
Solution:
<\/strong>Let fixed charges for the first 3 days = \u20b9x
and additional charges for each day = \u20b9y
Total period = 7 days
and amount charges = \u20b927
\u2234 x + (7 \u2013 3) x\u00a0y\u00a0\u00a0= 27
\u21d2\u00a0 x + 4y = 27
Hence x + 4y = 27<\/p>\nQuestion 11.
<\/strong>A number is 27 more than the number obtained by reversing its digits. If its unit\u2019s and ten\u2019s digits are x and y respectively, write the linear equation representing the above statement.
<\/strong>Solution:
<\/strong>Let the unit\u2019s digit = x
and tens digit = y
\u2234\u00a0 Number = x + 10y
By reversing the digits, units digit = y
and ten\u2019s digit = x
\u2234\u00a0 number = y + 10x
Now difference of these two numbers = 27 (x + 10y) \u2013 (y +10x) = 27
x + 10y \u2013 y \u2013 10x = 27
\u21d2\u00a0 -9x + 9y \u2013 27 = 0
\u21d2 x-_y + 3 = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Dividing by -9)
Hence equation is x \u2013 y + 3 = 0<\/p>\nQuestion 12.
<\/strong>The sum of a two-digit number and the number obtained by reversing the order of its digits is 121. If units and ten digits of the number are x and y respectively, then write the linear equation representing the above statement.
<\/strong>Solution:
<\/strong>Let unit digit = x
and tens digit = y
\u2234 Number = x + 10y
By reversing the digits,
units digit = y
and tens digit = x
\u2234 Number =y+ 10x
Now sum of these two numbers = 121
\u2234 x + 10y + y + 10x = 121
\u21d2\u00a0 1 lx + 11y = 121
\u21d2\u00a0 x + y = 11\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Dividing by 11)
\u2234\u00a0 x + y = 11<\/p>\nQuestion 13.
<\/strong>Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.
<\/strong>Solution:
<\/strong>2x + y = 6
\u21d2\u00a0 y = 6 \u2013 2x
If x = 0, then y = 6- 2 x 0 = 6 \u2013 0 = 6
If x = 2, then y = 6- 2 x 2 = 6- 4 = 2
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q13.1.png\")
Now plot the points (0, 6) and (2, 2) on the graph and join them to get a line that intersects the x-axis at (3, 0) and the y-axis at (0,6)
Now coordinates if vertices of the shaded portion are (6, 0) (0, 0) and (3, 0) Now the area of the shaded region.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q13.2.png\")
<\/p>\nQuestion 14.
<\/strong>Draw the graph of the equation\u00a0\u00a0x<\/span>3<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0+\u00a0\u00a0y<\/span>4<\/span><\/span><\/span><\/span><\/span><\/span> \u00a0= 1 Also find the area of the triangle formed by the line and the coordinate axes.
<\/strong>Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q14.1.png\")
Now plot the points (3, 0) and (0, 4) and join them to get a line with interests x-axis at A (3, 0) and y-axis at B (0, 4)
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q14.2.png\")
<\/p>\nQuestion 15.
<\/strong>Draw the graph of y = | x |
Solution:
<\/strong>y = | x |
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q15.1.png\")
\u21d2\u00a0 \u00a0y = x\u00a0 \u00a0 \u00a0 \u00a0[\u2235 | x |<\/span>=x]
\u2234\u00a0 Now taking z points.
<\/span>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q15.2.png\")
Now plot the points (1, 1) (2, 2), and (3, 3) and join them to get a graph of a line.<\/p>\nQuestion 16.
<\/strong>Draw the graph of y = | x | + 2<\/strong>
Solution:
<\/strong>y \u2013 | x | + 2
\u21d2\u00a0 y = x + 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0[| x | = x]
If x = 0, then y = 0 + 2 = 2
If x = 1, then y = 1+2 = 3
If x = 2, then y = 2 + 2 = 4
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q16.1.png\")
Now plot the points (0, 2), (1, 3), and (2, 4) on the graph and join them to get a line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q16.2.png\")
<\/p>\nQuestion 17.
<\/strong>Draw the graphs of the following linear equation on the same graph paper.<\/strong>
2x + 3y = 12, x -y = 1<\/strong>
Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis. Also, find the area of the triangle.
<\/strong>Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q17.1.png\")
Now plot the points (6, 0) (0, 4) on the graph to get a line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q17.2.png\")
Now plot the points (1, 0) and (2, 1) on the graph to get another line.
Area of the triangle FEB so formed,
=\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0FB x FL =\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0x 5 x 3
=\u00a015<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span>
= 7.5 sq. units
co-ordinates of E, F, B are E (3, 2), (0, -1) and (0, 4)<\/p>\nQuestion 18.
<\/strong>\u00a0Draw the graphs of the linear equations 4x \u2013 3y + 4 = 0 and 4x + 3y \u2013 20 = 0. Find the area bounded by these lines and the x-axis.<\/strong>
Solution:<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q18.1.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q18.2.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q18.3.png\")
Now plot the points (5, 0) and (2, 4)and join them to get a line we see that the \u0394ABC is formed by bounding their line with the x-axis.<\/p>\nQuestion 19.
<\/strong>The path of a train A is given by the equation 3x + 4y \u2013 12 = 0 and the path of another train B is given by the equation 6jc + 8y \u2013 48 = 0. Represent this situation graphically.
<\/strong>Solution:
<\/strong>Path of the train A = 3x + 4y \u2013 12 = 0
Path of the train B = 6x + 8y \u2013 48 = 0
Now, 3x + 4y \u2013 12 = 0
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q19.1.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q19.2.png\")
Now plot the points (4, 0) and (0, 3) on the graph and join them to get a line and 6x + 8y \u2013 48 = 0
\u21d2\u00a0 6x = 48 \u2013 8y
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q19.3.png\")
Now plot the points (0, 6) and (4, 3) on the graph and join them to get another line.<\/p>\nQuestion 20.
<\/strong>Ravish tells his daughter Aarushi, \u201cSeven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be\u201d. If the present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.<\/strong>
Solution:
<\/strong>The present age of Aarushi = x years
and age of Ravish = y years
7 years ago,
age of Aarushi = x \u2013 7
years and age of Ravish =y-7 years
\u2234 y- 7 = 7 (X \u2013 7)
\u21d2\u00a0 y \u2013 7 = 7x \u2013 49
\u21d2\u00a0 7x \u2013 y = -7 + 49\u00a0 = 42
7x \u2013 y = 42
\u21d2\u00a0 y = 7x \u2013 42
If x = 6, then
y = 7 x 6 \u2013 42 = 42 \u2013 42 = 0,
If x = 7, then
y\u00a0= 7 x 7 \u2013 42 \u2013 49 \u2013 42 = -7
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q20.1.png\")
Plot the points (6, 0) (7, -7) on the graph and join them.
After 3 years,
age of Aarushi = x + 3
and age of Ravish = y + 3
\u21d2\u00a0 y + 3 = 3(x + 3)
\u21d2 y + 3 = 3x + 9
\u21d2 y = 3x+ 9-3
\u21d2 y = 3x + 6
If x = -2, then y = 3 x (-2) + 6 =6-6=0
If x = 1, then y = 3 x (1) + 6 =3+6=9
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q20.2.png\")
Plot the points (1, 9), (-2, 0) on the graph Arundeep\u2019s Mathematics (R.D.) 9th and join them to get another line.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q20.3.png\")
<\/p>\nQuestion 21.
<\/strong>Aarushi was driving a car with a uniform speed of 60 km\/h. Draw the distance-time graph. From the graph, find the distance traveled by Aarushi in.<\/strong>
(i) 21<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0Hours\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>
(ii)\u00a01<\/span>2<\/span><\/span><\/span><\/span><\/span><\/span>\u00a0Hour
<\/strong>Solution:
<\/strong>Speed of car = 60 km \/ h.
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q21.1.png\")
Now plot the points (60, 1), (120, 2) are the graph and join them to get the graph of the line.
From the graph, we see that
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q21.2.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-9-Solutions-Chapter-7-Introduction-to-Euclid%E2%80%99s-Geometry-Ex-7.3-Q21.3.png\")
<\/p>\nThis is the complete blog of RD Sharma Class 9 Solutions Chapter 7 Exercise 7.3. To know more about the CBSE<\/a> Class 9 Maths exam, ask in the comments.\u00a0<\/p>\n
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-7-Exercise-7.3.jpg\")
<\/p>\n