Question

{\pmb{\sf{\underline{Understanding \: the \: Question...}}}}UnderstandingtheQuestion...​UnderstandingtheQuestion...​

★ 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{Given \: that...}}}}Giventhat...​Giventhat...​

★ 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{To \; find...}}}}Tofind...​Tofind...​

★ The third vertex of the traingle

{\pmb{\sf{\underline{Solution...}}}}Solution...​Solution...​

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

{\pmb{\sf{\underline{Assumptions...}}}}Assumptions...​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{Full \; Solution...}}}}FullSolution...​FullSolution...​

~ 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}}}}}}:⟹4−1−3−2​​​

~ Now OZ will be the perpendicular to XY

~ Henceforth, the slope of OZ be the following,

{\sf{:\implies \dfrac{b-2}{a-1}}}:⟹a−1b−2​

{\sf{:\implies \: \: \therefore \: \dfrac{b-2}{a-1} \: = \dfrac{3}{4}}}:⟹∴a−1b−2​=43​

~ Let us cross multiply.

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

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

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

{\sf{:\implies \dfrac{b+3}{a-4}}}:⟹a−4b+3​

~ Henceforth, the slope of XZ be the following,

{\sf{:\implies \dfrac{5-2}{-2-1}}}:⟹−2−15−2​

{\sf{:\implies \dfrac{3}{-3}}}:⟹−33​

{\sf{:\implies -1}}:⟹−1

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

{\sf{:\implies \dfrac{b+3}{a-4} \: = 1}}:⟹a−4b+3​=1

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

{\sf{:\implies a-b-7 \: = 0 \dots Eq. \: 2}}:⟹a−b−7=0…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}}:⟹3a−4b+5=0

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

~ Solving this we get the following,

{\sf{:\implies a \: = 33}}:⟹a=33

{\sf{:\implies b \: = 26}}:⟹b=26

Henceforth, Z(a,b) is (33,26)

Henceforth, the third vertex of (33,26)

{\pmb{\sf{\underline{Additional \; Knowledge...}}}}AdditionalKnowledge...​AdditionalKnowledge...​

\underline{\bigstar\:\textsf{Distance Formula\; :}}★Distance Formula:​

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

{\underline{\boxed{\sf{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}Distance=(x2​−x1​)2+(y2​−y1​)2​​​ ⠀

\underline{\bigstar\:\textsf{Section Formula\; :}}★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{\frak{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}(x,y)=(m+nmx2​+nx1​​m+nmy2​+ny1​​)​​ ⠀

\underline{\bigstar\:\textsf{Mid Point Formula\; :}}★Mid Point Formula:​

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

{\underline{\boxed{\frak{\quad \Bigg(\dfrac{x_1 + x_2}{2} \; or\; \dfrac{y_1 + y_2}{2} \Bigg)\quad}}}}(2x1​+x2​​or2y1​+y2​​)​​

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

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