In this page equation of line solution16 we are going to see solution of each problem with detailed explanation of the worksheet slope of the line.

(17) If x + 2 y = 7 and 2 x + y = 8 are the equations of the lines of two diameter of the circle, find the radius of the circle if the point (0,-2) lie on the circle.

**Solution:**

Two diameters are intersecting at the point E

x + 2 y = 7 --- (1)

2 x + y = 8 --- (2)

(1) x 2 = > 2 x + 4 y = 14

(1) - (2) 2 x + y = 8

(-) (-) (-)

--------------

3 y = 6

y = 6/3

y = 2

Substitute y = 2 in the first equation

x + 2 (2) = 7

x + 4 = 7

x = 7 - 4

x = 3

Therefore two diameters are intersecting at the point E (3 , 2)

Now we have to find the measurement of radius for that we have to find the distance between the points F(0 , -2) and E(3 , 2)

d = √(x₂ - x₁)² + (y₂ - y₁)²

= √(3 - 0)² + (2 - (-2))²

= √(3)² + (2 + 2)²

= √9 + 4²

= √9 + 16

= √25

= √5 x 5

= 5 units

(18) Find the equation of the straight line segment whose end points are the point of intersection of the straight lines 2 x – 3 y + 4 = 0, x – 2 y + 3 = 0 and the midpoint of the line joining the points (3 ,-2) and (-5 , 8).

**Solution:**

2 x – 3 y + 4 = 0 ----- (1)

x – 2 y + 3 = 0 ----- (2)

Now we have to find the point of intersection of those two lines

2 x – 3 y + 4 = 0

(2) x 2 = > 2 x - 4 y + 6 = 0

(-) (+) (-)

------------------

y - 2 = 0

y = 2

Substitute y = 2 in the first equation

2 x - 3 (2) + 4 = 0

2 x - 6 + 4 = 0

2 x - 2 = 0

2 x = 2

x = 2/2

x = 1

Now we have to find the midpoint of the line segment joining the points

(3 ,-2) and (-5 , 8)

Midpoint = (x₁ + x₂)/2 , (y₁ + y₂)/2

= (3 - 5)/2 , (- 2 + 8)/2

= -2/2 , 6/2

= (-1 , 3)

(19) If the isosceles triangle PQR, PQ = PR. The base QR lies on the axis, P lies on the y-axis and 2 x – 3 y + 9 =0 is the equation of PQ. Find the equation of PQ. Find the equation of the straight line along PR.

To find the point point P which lies on the y-axis we have to put 0 instead of x in the given equation PQ

2(0) - 3 y = - 9

- 3 y = -9

y = (-9)/(-3)

y = 3

Therefore the point P is (0,3)

To find the point point Q which lies on the x-axis we have to put 0 instead of y in the given equation QR

2 x - 3 (0) = - 9

2 x - 0 = -9

2 x = - 9

x = -9/2

Therefore the point Q is (-9/2 , 0)

The length of the sides PQ and PR are same.So the point R is (9/2,0)

Equation of PR

P (3 , 0) R (9/2 , 0)

(y-y₁)/(y₂ - y₁) = (x-x₁)/(x₂ - x₁)

(y - 3)/(0 - 3) = [x - (9/2)]/[(9/2) - 3)]

(y - 3)/(- 3) = [2x - 9]/[(9-6)/2]

(y - 3)/(- 3) = [2x - 9]/(3/2)

(y - 3)/(- 3) = 2[2x - 9]/3

-y + 3 = 4 x - 18

4 x + y - 18 - 3 = 0

4 x + y - 21 = 0

equation of line solution16 equation of line solution16 equation of line solution16