Question

Two vertices of a triangle are (4,-3) and (-2,5). If the orthocentre of the triangle is at (1,2), then the third vertex is

Answer 1

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

★ 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).

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

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

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

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

★ The third vertex of the traingle

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

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

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

● 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)

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

~ 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)

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

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

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

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

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

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.

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

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

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

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

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

Answer 2

Two vertices of a triangle are A(4,-3) and B(-2,5). The orthocentre of the triangle is at O(1,2). Orthocenter of a triangle is the point of intersection of the altitudes drawn to each side of the triangle.

Let C(x, y) be the third vertex.

The side vectors of the triangle are,

• $\vec{AB}=-6\,\hat i+8\,\hat j$

• $\vec{BC}=(x+2)\,\hat i+(y-5)\,\hat j$

• $\vec{AC}=(x-4)\,\hat i+(y+3)\,\hat j$

The vectors joining the orthocenter with each vertex are,

• $\vec{OA}=3\,\hat i-5\,\hat j$

• $\vec{OB}=-3\,\hat i+3\,\hat j$

• $\vec{OC}=(x-1)\,\hat i+(y-2)\,\hat j$

The vector $\vec{OA}$ belongs to the altitude drawn to the line BC, so the vectors

$\longrightarrow\vec{OA}\cdot\vec{BC}=0$

$\longrightarrow(3\,\hat i-5\,\hat j)\cdot\left((x+2)\,\hat i+(y-5)\,\hat j\right)=0$

$\longrightarrow3(x+2)-5(y-5)=0$

$\longrightarrow3x-5y+31=0$

$\longrightarrow y=\dfrac{3x+31}{5}\quad\quad\dots(1)$

Similarly the vectors $\vec{OB}$ and $\vec{AC}$ are perpendicular, hence,

$\longrightarrow\vec{OB}\cdot\vec{AC}=0$

$\longrightarrow(-3\,\hat i+3\,\hat j)\cdot\left((x-4)\,\hat i+(y+3)\,\hat j\right)=0$

$\longrightarrow-3(x-4)+3(y+3)=0$

$\longrightarrow x-y-7=0$

Putting value of y from (1),

$\longrightarrow x-\dfrac{3x+31}{5}-7=0$

$\longrightarrow 2x-66=0$

$\longrightarrow x=33$

Then (1) becomes,

$\longrightarrow y=\dfrac{3(33)+31}{5}$

$\longrightarrow y=26$

Check for equating dot product of $\vec{OC}$ and $\vec{AB}$ to zero.

$\longrightarrow\vec{OC}\cdot\vec{AB}=0$

$\longrightarrow\left((x-1)\,\hat i+(y-2)\,\hat j\right)\cdot\left(-6\,\hat i+8\,\hat j\right)=0$

$\longrightarrow-6(x-1)+8(y-2)=0$

$\longrightarrow3x-4y+5=0$

Putting values of x and y,

$\longrightarrow3(33)-4(26)+5=0$

$\longrightarrow0=0$

Hence the third vertex is (33, 26).