Are you Ricky? The same question was posted a day or two back, which I answered the first part just now.

In any case, once k is known to be 0 or 6, there are two possible coordinates of C.

Let's call them C₁(0,2) and C₂(6,8)

If you try to sketch the two triangles, both have a common side AB, but ABC₂ is obtuse and
ABC₁ is acute.

We need to find the height which is perpendicular to the base.

The idea is to draw a perpendicular line to AB from C₁ since it lies below AB.

We find the coordinates of the foot of the perpendicular from AB to C₁ (call it D) by solving for the intersection between the two lines. Then use the coordinates to find the length C₁D to get our height.

(You can use C₂ as well but AB has to be extended since it lies beyond AB as it is the rightmost point)

To solve,

Find the equation of AB first.

Gradient = (5 - 9)/(3 - (-5)) = -4/8 = -½

Sub A(3,5) and m = -½ into y - y1 = m(x - x1),

y - 5 = -½(x - 3)
y - 5 = -½x + 1½
y = -½x + 6½
(or y = -0.5x + 6.5)

Gradient of the perpendicular line(the 'height') = -1/-½
= 2

Since C₁(0,2) lies on the line and the y-axis(x = 0), we know it is the y-intercept.

Equation of the perpendicular line is :

y = 2x + 2

Find the intersection point between the two lines, D₁

-0.5x + 6.5 = 2x + 2
2.5x = 4.5
x = 4.5 / 2.5
x = 1.8

y = 2(1.8) + 2 = 5.6

So D₁(1.8,5.6)


height = C₁D₁

= √[(1.8 - 0)² + (5.6 - 2)²]
= √(1.8² + 3.6²)
= √16.2
=√81/5

= 9/√5 or 9√5 / 5

If you choose to use C₂ instead, then:

Sub m = 2 and C₂(6,8) into y - y1 = m(x - x1),

y - 8 = 2(x - 6)
y = 2x - 4
(Equation of the other perpendicular line)

Again, find the intersection point between the two lines, D₂

-0.5x + 6.5 = 2x - 4
2.5x = 10.5
x = 10.5 / 2.5
x = 4.2

y = 2(4.2) - 4 = 4.4

So D₂(4.2,4.4)


height = C₂D₂

= √[(6 - 4.2)² + (8 - 4.4)²]
= √(1.8² + 3.6²)
= √16.2
= √(81/5)
= 9/√5 or 9√5 / 5