Find the coordinates of the vertices of a triangle if given the coordinates of one of its vertices B (6; 14)

| June 17, 2021 | education

Find the coordinates of the vertices of a triangle if given the coordinates of one of its vertices B (6; 14) and the equations of its height: x + 4y-9 = 0 and the bisector 4x + 7y-12 = 0 drawn from one vertex.

Suppose that this triangle has vertices A, B and C, in which the coordinates of the vertex B are given (6; 14). Let the height (with the equation x + 4 * y – 9 = 0) and the bisector (with the equation 4 * x + 7 y – 12 = 0) are drawn from the vertex A.
Obviously, if we solve together (as a system of linear equations) the equations of height and bisector, then we find the coordinates of the vertex A. Using the equation of height, we find: x = –4 * y + 9. Substitute this expression into the equation of the bisector. Then 4 * (–4 * y + 9) + 7 y – 12 = 0 or -16 * y + 7 * y = 12 – 36, whence y = -24 / (-9) = 8/3. Therefore, x = –4 * (8/3) + 9 = (-32 + 27) / 3 = -5/3. Hence, the coordinates of the vertex A are (-5/3; 8/3).
In order to find the coordinates of the vertex C, we first compose the equation for the side BC of the triangle ABC. Let the sought equation of the line BC have the form y = k * x + b. Since the height drawn through the vertex A is perpendicular to BC, then bringing the equation of height to the form y = -¼ * x + 9/4, we will use the condition of perpendicularity of lines: -¼ * k = -1. It is clear that k = 4. Point B lies on the straight line BC, therefore, the equality 14 = 4 * 6 + b is true. We have: b = 14 + 24 = -10. This means that the desired equation of the line BC has the form y = 4 * x – 10.
Now let’s make the equation of the line AB. To do this, we will use an equation passing through the given two points. We have: (x – (-5/3)) / (6 – (-5/3)) = (y – 8/3) / (14 – 8/3) or 34 * x – 23 * y + 118 = 0, whence y = (34/23) * x + 118/23.
We rewrite the bisector equation in the form y = – (1/7) * x + 12/7. Suppose that the bisector of angle A intersects the side BC at point D. Find the tangent of the angle ∠BAD. Let’s calculate: tg∠BAD = ((-1/7) – (34/23)) / (1 + (-1/7) * (34/23)) = -261/127.
Let’s compose the equation of the straight line AC. Let it have the form y = k * x + b. According to the terms of the assignment, ∠BAD = ∠DАС. Then tg∠BAD = tg∠DАС. We have: tg∠DАС = (k – (-1/7)) / (1 + k * (-1/7)) = (7 * k + 1) / (7 – k). Then (7 * k + 1) / (7 – k) = -261/127 or 127 * (7 * k + 1) = -261 * (7 – k), whence k = (-261 * 7 – 127) / (127 * 7 – 261) = -1954/628 = -977/314. Point A lies on the straight line AC. Therefore, the equality 8/3 = -977/314 * (-5/3) + b is true. We have: b = (8/3) – (977 * 5) / (314 * 3) = -791/314. This means that the desired equation of the line AC has the form y = (-977/314) * x – 791/314.
We find the coordinates of the point C by solving together the equations of the straight lines AC and BC. We have: 4 * x – 10 = (-977/314) * x – 791/314, whence x = (10 – 791/314) / (4 + 977/314) = 81/77. Calculate: y = 4 * (81/77) – 10 = -446/77. Therefore, the coordinates of vertex C are (81/77; -446/77).
Answer: A (-5/3; 8/3), B (6; 14) and C (81/77; -446/77).



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