## PROBLEMS ON INTERSECTING LINES

Example 1 :

If x + 2y = 7 and 2x + 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) lies on the circle.

Solution :

Two diameters are intersecting at the point E

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

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

2(1) - (2) :

2(x + 2y) - (2x + y) =  14 - 8

2x + 4y - 2x - y = 6

3y = 6

y = 2

Substitute y = 2 in (1).

x + 2(2) = 7

x + 4 = 7

x = 7 - 4

x = 3

Diameters are intersecting at the point E (3, 2).

So, center of the circle is E(3, 2).

To find radius, find the distance between the points F and E.

d = √[(x- x1)+ (y- y1)2]

Substitute (x1, y1) = E(3, 2) and (x2, y2) = F(0, -2).

= √[(0 - 3)+ (-2 - 2)2]

= √(9 + 16)

= √25

= 5

Radius of the circle is 5 units.

Example 2 :

Find the equation of the straight line segment whose end points are the point of intersection of the straight lines

2x – 3y + 4 = 0, x – 2y + 3 = 0

and the midpoint of the line joining (3, -2) and (-5, 8).

Solution :

2x – 3y + 4 = 0 ----(1)

x – 2y + 3 = 0 ----(2)

Let A be the point of intersection of (1) and (2) and B be the midpoint of the line joining (3, -2) and (-5, 8).

Target : Find the equation of the line AB.

(1) - 2(2) :

2x - 3y + 4 - 2(x - 2y + 3) = 0

2x - 3y + 4 - 2x + 4y - 6 = 0

y - 2 = 0

y = 2

Substitute y = 2 in (1).

2x - 3(2) + 4 = 0

2x - 6 + 4 = 0

2x - 2 = 0

x = 1

Point of intersection of the lines (1) and (2) is A(1, 2).

Find the midpoint of line joining (3, -2) and (-5, 8).

Midpoint = ((x+ x2)/2, (y+ y2)/2)

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

= (-2/2, 6/2)

= B(-1, 3)

Find the slope of the line AB :

m = (y2 - y1)/(x2 - x1)

Substitute (x1, y1) = A(1, 2) and (x2, y2) = B(-1, 3).

m = (3 - 2)/(-1 - 1)

= 1/(-2)

= -1/2

Equation of the line AB :

y - y1 = m(x - x1)

Substitute (x1, y1) = A(1, 2) and m = -1/2.

y - 2 = (-1/2)(x - 1)

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

2y - 4 = -x + 1

x + 2y - 5 = 0

Example 3 :

If the isosceles triangle PQR, PQ = PR. The base QR lies on the axis, P lies on the y-axis and

2x – 3y + 9 = 0

is the equation of PQ. Find the equation of PQ. Find the equation of the straight line along PR.

Solution :

To find the point point P which lies on the y-axis, substitute x = 0.

2(0) - 3y = -9

-3y = -9

y = 3

The point P is (0, 3)

To find the point point Q which lies on the x-axis, substitute y = 0.

2x - 3(0) = -9

2x - 0 = -9

2x = -9

x = -9/2

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 :

(y-y1)/(y2-y1) = (x-x1)/(x2-x1)

Substitute (x1, y1) = P(3, 0) and (x2, y2) = R(9/2, 0).

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

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

-(y - 3) = 2(2x - 9)

- y + 3 = 4x - 18

4x + y - 21 = 0

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