Question

5. Find the equations of perpendicular
bisectors of sides of the triangle whose
vertices are P(-1,8), 2(4,-2) and R(-5,-3).​

Answer 1

Step-by-step explanation:

$\huge\bf\underline{Given:}$

=> Vertices of sides of triangle are (-1,8),Q(4,-2),R(-5,-3).

$\huge\bf\underline{To \: Find:}$

=>The equations of perpendicular bisectors of sides of the triangle.

$\sf\overbrace{ = > Understanding \: the \: concept:}$

=> Let A,B and C be the midpoints of sides PQ,QR and PR respectively of ∆PQR.

=> A is the midpoint of side PQ.

$A=> ( \frac{ - 1 + 4}{2</p><p>} </p><p>, \frac{8 - 2}{2} )$

$A=> \frac{3}{2} </p><p>,2$

$\sf{Slope \: of \: side \: PQ = \frac{ - 2 - 8}{4 - ( - 1)} }$

$A=> \frac{ - 10}{5} = > - 2$

So, Slope of perpendicular bisectors of PQ is 1/2 and it passes through 3/2,3.

=> Equation of the perpendicular bisector of side PQ is,

$\sf{y - 3 = \dfrac{1}{2} (x - \dfrac{3}{2} )}$

$\sf{}y - 3 = \dfrac{1}{2} ( \dfrac{2x - 3}{2})$

$\sf{4(y - 3) = 2x - 3}$

$\sf{}4y - 12 = 2x - 3$

$\sf{2x - 4y + 9 = 0}$

B is the midpoint of side QR.

$\sf{b = > ( \dfrac{4 - 5}{2} </p><p>, \dfrac{ - 2 - 3}{2} )}$

$\sf{b = > \dfrac{ - 1}{ \: 2} </p><p>, \dfrac{ - 5}{2} }$

Slope of perpendicular side QR,

$\sf{ = > \dfrac{ - 3 - ( - 2)}{ - 5 - 4} } = > \dfrac{ - 1}{ - 9} = > \dfrac{1}{9}$

Slope of perpendicular bisector of QR is -9 and it passes through -1/2, -5/2.

=> Equation of perpendicular bisector of side QR is,

$\sf{y - (\frac{ - 5}{2} ) = - 9(x - ( \frac{ - 1}{2} )})$

$\sf{ \dfrac{2y + 5}{2} = - 9( \dfrac{2x + 1}{2} )}$

$\sf{2y + 5 = - 18x - 9}$

$\sf{18x + 2y + 14 = 0}$

$\sf{9x + y + 7 = 0}$

C is the midpoint of side PR.

$\sf{C = > \dfrac{ - 1 - 5}{2} </p><p>, \dfrac{8 - 3}{2} }$

$\sf{ = > - 3</p><p>, \dfrac{5}{2} }$

Slope of side PR,

$\sf{ = > \dfrac{ - 3 - 8}{ - 5 - ( - 1)} } = > \dfrac{ - 11}{ - 4}$

$\sf{ = > \dfrac{11}{4} }$

Slope of perpendicular bisector of PR is -4/11 and it passes through (-3,-5/2).

=> Equation of the perpendicular bisectors of side PR is,

$\sf{y - \dfrac{ 5}{2} = \dfrac{ - 4}{11} (x + 3)}$

$\sf{11( \dfrac{2y - 5}{2} = - 8( x + 3) }$

$\sf{22y - 5 = - 8x - 24}$

$\sf{8x + 22y - 31 = 0}$

Therefore, We found all the equations!!.