Question

The coordinates of points A, B and C are

(0, 4), (9, 7) and (8,0) respectively.
The points L, M, N are on the sides BC, CA and AB respectively of the triangle ABC such that AL
is perpendicular to BC and BM is perpendicular
to CA.
The equation of the line AL is 7y + x = 28.
(i)
Find the gradient of the line AC and
hence the gradient of the line BM.
(ii)
Find the equation of the line BM.
iii)
Find the coordinates of the point P
where AL and BM intersect.

Answer 1

Part (i)

We can use the slope formula on the points A(0,4) and C(8,0)

`m = (y_2-y_1)/(x_2-x_1)\\\\ m = (0-4)/(8-0)\\\\ m = (-4)/(8)\\\\ m = -(1)/(2)`

The slope of line AC is -1/2.

Line BM is perpendicular to AC. This indicates we'll apply the negative reciprocal to -1/2 to end up with 2.

The slope of line BM is 2.

Note that the perpendicular slopes multiply to -1. This is true of any perpendicular pair of slopes as long as neither line is vertical nor horizontal.

Answers:

• Slope of line AC = -1/2
• Slope of line BM = 2

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Part (ii)

The slope of
`\text{line }\text{B}\text{M}` was m = 2 from the previous part above.

We'll use the coordinates of point B(9,7) along with this slope value to find the equation of
`\text{line }\text{B}\text{M}`.

`y - y_1 = m(x - x_1)\\\\ y - 7 = 2(x - 9)\\\\ y - 7 = 2x - 18\\\\ y = 2x - 18+7\\\\ y = 2x - 11\\\\`

Answer: y = 2x-11

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Part (iii)

The equations for lines
`\text{A}\text{L}` and
`\text{B}\text{M}` are
`7y+x = 28` and
`y = 2x-11` in that order. Solving this system of equations will produce the location of point P, where the lines intersect.

Let's apply substitution to solve for x.

`7y+x = 28\\\\ 7( y ) + x = 28\\\\ 7( 2x-11 ) + x = 28 \ \text{ ... y replaced with 2x-11}\\\\ 14x-77+x = 28\\\\ 15x = 28+77\\\\ 15x = 105 \\\\ x = 105/15\\\\ x = 7\\\\`

Then use this to find the y coordinate.

`y = 2x-11\\\\ y = 2(7)-11\\\\ y = 14-11\\\\ y = 3`

The location of point P is (7,3). This is the orthocenter of the triangle where the altitudes intersect.

Answer: (7,3)