The vertex A of a triangle ABC is the point (-2, 3) whereas the line perpendicular to the sides AB and AC are x – y – 4 = 0 and 2x – y – 5 = 0 respectively. The right bisectors of sides meet at P(3/2 , 5/2) . Then the equation of the median of side BC isa)5x + 2y = 10b)5x – 2y = 16c)2x – 5y = 10d)none of theseCorrect answer is option 'D'. Can you explain this answer?

Question

The vertex A of a triangle ABC is the point (-2, 3) whereas the line perpendicular to the sides AB and AC are x &ndash; y &ndash; 4 = 0 and 2x &ndash; y &ndash; 5 = 0 respectively. The right bisectors of sides meet at P(3/2 , 5/2) . Then the equation of the median of side BC isa)5x + 2y = 10b)5x &ndash; 2y = 16c)2x &ndash; 5y = 10d)none of theseCorrect answer is option 'D'. Can you explain this answer?

Answer

We need to find the equations of the perpendicular lines to sides AB and AC passing through A.

First, let's find the slope of side AB.

slope(AB) = (yB - yA) / (xB - xA)

We don't have the coordinates of point B, but we know that side AB is perpendicular to a line with equation x - 2y = 5.

The slope of this line is:

slope(perpendicular to AB) = -1 / slope(AB)

So,

-1 / slope(AB) = -2

slope(AB) = 1/2

Now we can find the equation of the line perpendicular to AB passing through A:

y - 3 = -2(x + 2)

y = -2x - 1

Similarly, let's find the slope of side AC.

slope(AC) = (yC - yA) / (xC - xA)

We don't have the coordinates of point C, but we know that side AC is perpendicular to a line with equation 3x - y = 7.

The slope of this line is:

slope(perpendicular to AC) = -1 / slope(AC)

So,

-1 / slope(AC) = 3

slope(AC) = -1/3

Now we can find the equation of the line perpendicular to AC passing through A:

y - 3 = (-1/3)(x + 2)

y = (-1/3)x + 7/3

Therefore, the equations of the perpendicular lines to sides AB and AC passing through A are:

y = -2x - 1 (perpendicular to AB)

y = (-1/3)x + 7/3 (perpendicular to AC)

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