Question

CIRCUMCLNIR
Two vertices of a triangle are (4,-3) and
(-2,5), if the orthocentre of the triangle is
(1,2) then its third vertex is​

Answer 1

Answer:

4-(-3)=+7

+7(-2)+5=17

17-1-2=14

answer is14

Answer 2

${\pmb{\sf{\underline \red{Understanding \: the \: Question...}}}}$

★ This question says that we have to find out the third vertex of the traingle whose two vertices are given as (4,-3) and (-2,5) and the orthocentre of the triangle is at (1,2).

${\pmb{\sf{\underline \red{Given \: that...}}}}$

★ The traingle whose two vertices are given as (4,-3) and (-2,5)

★ The orthocentre of the triangle is at (1,2).

${\pmb{\sf{\underline \red{To \; find...}}}}$

★ The third vertex of the traingle

${\pmb{\sf{\underline \red{Solution...}}}}$

★ The third vertex of the traingle = (33,26)

${\pmb{\sf{\underline \red{Assumptions...}}}}$

● Let the first vertex of triangle be X(4,-3)

● Let the second vertex of triangle be Y(-2,5)

● Let the orthocentre be O(1,2)

● Let the third vertex of triangle be Z(a,b)

${\pmb{\sf{\underline \red{Full \; Solution...}}}}$

~ As we have to find out the third vertex of the traingle whose two vertices are given as (4,-3) and (-2,5) and the orthocentre of the triangle is at (1,2).

~ Henceforth, slope coming from X is given as the following,

${\small{\underline{\boxed{\sf{:\implies \dfrac{-3-2}{4-1}}}}}}$

~ Now OZ will be the perpendicular to XY

~ Henceforth, the slope of OZ be the following,

${\sf{:\implies \dfrac{b-2}{a-1}}}$

${\sf{:\implies \: \: \therefore \: \dfrac{b-2}{a-1} \: = \dfrac{3}{4}}}$

~ Let us cross multiply.

${\sf{:\implies 4b \: -8 \: = 3a \: - 3}}$

${\sf{:\implies 3a \: - 4b \: + 5 \: = 0 \dots Eq. \: 1}}$

~ Now by the similar way the slope of XZ be the following,

${\sf{:\implies \dfrac{b+3}{a-4}}}$

~ Henceforth, the slope of XZ be the following,

${\sf{:\implies \dfrac{5-2}{-2-1}}}$

${\sf{:\implies \dfrac{3}{-3}}}$

${\sf{:\implies -1}}$

~ Since, as the OY is perpendicular to XZ henceforth,

${\sf{:\implies \dfrac{b+3}{a-4} \: = 1}}$

${\sf{:\implies b+3 \: = a-4}}$

${\sf{:\implies a-b-7 \: = 0 \dots Eq. \: 2}}$

~ Now at last we have to use Eq. 1 and Eq. 2 to find our final result.

${\sf{:\implies 3a \: - 4b \: + 5 \: = 0}}$

${\sf{:\implies a-b-7 \: = 0}}$

~ Solving this we get the following,

${\sf{:\implies a \: = 33}}$

${\sf{:\implies b \: = 26}}$

Henceforth, Z(a,b) is (33,26)

Henceforth, the third vertex of (33,26)

${\pmb{\sf{\underline \red{Additional \; Knowledge...}}}}$

$\underline{\bigstar\:\textsf{Distance Formula\; :}}$

• Distance formula is used to find the distance between two given points.

${\underline{\boxed{\sf \pink{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

$\underline{\bigstar\:\textsf{Section Formula\; :}}$

• Section Formula is used to find the co ordinates of the point(Q) which divides the line segment joining the points (B) and (C) internally or externally.

${\underline{\boxed{\sf \pink{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

$\underline{\bigstar\:\textsf{Mid Point Formula\; :}}$

• Mid Point formula is used to find the mid points on any line.

${\underline{\boxed{\sf \pink{\quad \Bigg(\dfrac{x_1 + x_2}{2} \; or\; \dfrac{y_1 + y_2}{2} \Bigg)\quad}}}}$

And Orthocentre of a triangle is the point of intersection of it's three altitudes/height.