**Coordinate of the Foot of the Perpendicular From the Point to the Line : **

Here we are going to see how to find the coordinate of foot of perpendicular from the point to the line.

**Question 16 :**

Find the coordinates of the foot of the perpendicular from the origin on the straight line 3x + 2 y = 13

**Solution :**

Equation of the line joining the points (0, 0) and (a, b)

Slope of the line joining the points (0, 0) and (a, b)

m = (b - 0)/(a - 0)

m = b/a

Slope of the line 3x + 2y = 13

m = -3/2

Product of slopes = -1

(b/a) ⋅ (-3/2) = -1

3b/2a = 1

3b = 2a

b = 2a/3

The line 3x + 2y = 13 passing through the point (a, b)

3a + 2(2a/3) = 13

3a + (4a/3) = 13

(9a + 4a)/3 = 13

13a/3 = 13

a = 3

b = 2(3)/3

b = 2

Hence the foot of perpendicular is (3, 2).

**Question 17 :**

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

**Solution :**

To find the center of the circle we need to find the point of intersection of two given lines

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

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

(1) - (2)

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

2x
+ y = 8

(-) (-) (-)

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

3y = 6

y = 2

Substitute y = 2 in the first equation

x + 2(2) = 7

x + 4 = 7

x = 3

the point of intersection is (3,2)

Distance between the points (3,2) and (0,-2)

= √(x₂ - x₁)² + (y₂ - y₁)²

= √(0 - 3)² + (-2 - 2)²

= √(- 3)² + (-4)²

= √9 + 16

= √25

= 5 units

After having gone through the stuff given above, we hope that the students would have understood, how to find the coordinate of the foot of the perpendicular from the point to the line.

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