RD Chapter 7 Introduction to Euclid s Geometry Ex 7.3 Solutions
Question - 1 : - Draw the graph of each of the following linear equations in two variables.
Answer - 1 : -
(i)x + y = 4
x = 4 – 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 ro get the graph of the given equation
(ii)x – y = 2
x = 2 +y
If y = 0, then x = 2 and if y = 1,
Then x = 2 + 1 = 3
Now plot the points (2, 0) and (3, 1) on the graph and join them to get the graph of the equation.
(iii) -x+y = 6 ⇒ y = 6+x
If x = 0, then y = 6 + 0 = 6
If x = -1, then y = 6 – 1 = 5
Now plot the points (0, 6) and (-1, 5) on the graph and join them to get a graph of the line.
(iv) y = 2x
If x = 0, then y = 2 x 0 = 0
If x = 1, then y = 2 x 1 = 2
Now plot the points (0, 0) and (1, 2) on the graph and join them to get the graph of the line.
Now plot the points (5, 0) and (0, 3) on the graph and join them to get the graph of the line.
Now plot the points (4, 0) and (2, -3) on the graph and join them to get the graph of the line.
Now plot the points (-1, 2) and (2, 3) on the graph and join then to get the graph of the line.
Now plot the points (1, 0) and (-1, 1) on the graph and join then to get the graph of the line.
Question - 2 : - Give the equations of two lines passing through (3, 12). How many more such lines are there, and why ?
Answer - 2 : -
∵ Points (3, 12) lies on the lines passing through the points
∴ Solutions is x = 3,y- 12
∴ Possible equation can be
x + y = 15
-x+y = 9
4x-y = 0
3x – y + 3 = 0
Question - 3 : - A three-wheeler scooter charges ₹15 for first kilometer and ₹8 each for every subsequent kilometer. For a distance of x km, an amount of ₹y is paid. Write the linear equation representing the above information.
Answer - 3 : -
Charges for the first kilometer = ₹15
Charges for next 1 km = ₹8
Distance = x km
and total amount = ₹y
∴ Linear equation will be,
15 + (x- 1) x 8 =y
⇒ 15 + 8x – 8 = y
⇒ 7 + 8x = y
∴ y = 8x + 7
Question - 4 : - Plot the points (3, 5) and (-1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).
Answer - 4 : -
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.
Question - 5 : - From the choices given below, choose the equation whose graph is given in figure.
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x
Answer - 5 : -
From the graph, we see that Points (-1, 1) and (1, -1) be on the graph of the line these will satisfy the equation of the line
∴ -x = y ⇒ x+ y = 0
i.e, required equation
∵ x + y = 0 is the graph of the equation.
Question - 6 : - From the choices given below, choose the equation whose graph is given in figure.
(i) y = x + 2
(ii) y = x – 2
(ii) y = -x + 2
(iv) x + 2y = 6
Answer - 6 : -
From the graph
Points (-1,3) and (2, 0) lie on the graph of the line
Now there points, by observation, satisfy the equation y= -x+2
∴ Required equation is y = -x + 2 whose graph is given.
Question - 7 : - If the point (2, -2) lies on the graph of the linear equation 5x + ky =4, find the value of k.
Answer - 7 : -
∵ Point (2, -2) lies on the graph of the linear equation 5x + ky = 4
∴ It will satisfy it
∴ Now substituting the values of x and y 5 x 2 + k (-2) = 4
⇒ 10 – 2k = 4 ⇒ -2k = 4 – 10 = -6 -6
⇒ k= 
=3 Hence k = 3
Question - 8 : - Draw the graph of the equation 2x + 3p = 12. From the graph find the co-ordinates of the point.
(i) whose y -coordinates is 3
(ii) whose x-coordinates is -3
Answer - 8 : -
Plot the points (6, 0) and (0, 4) on the graph and join them to get the graph if the line.
(i) If y = 3, then draw perpendicular from y = 3 to the line, which get meets it at P then x-coordinate of p will be
∴ coordinates of P are ( 32 ,3)
(ii) If x = -3, draw perpendicular from x = -3 to the line, which meets it Q.
The y coordinates of Q will be y = 6
∴ co-ordinates of Q are (-3, 6)
Question - 9 : - Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinates axes:
(i) 6x – 3y = 12
(ii) -x + 4y = 8
(iii) 2x + y = 6
(iv) 3x + 2y + 6 = 0
Answer - 9 : -
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)
Answer - 10 : -
Let fixed charges for first 3 days = ₹x
and additional charges for each day = ₹y
Total period = 7 days
and amount charges = ₹27
∴ x + (7 – 3) x y = 27
⇒ x + 4y = 27
Hence x + 4y = 27