Two vertices of a triangle are (4, -3) and (- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the vertices of the triangle be $A(h, k)$,$B(4, -3)$,and $C(-2, 5)$. Let the orthocentre be $O(1, 2)$.$1$. The slope of $BC$ is $m_{BC} = \fra

Vedclass · Accepted Answer

$(33, 26)$

Vedclass · Answer

(B) Let the vertices of the triangle be $A(h, k)$,$B(4, -3)$,and $C(-2, 5)$. Let the orthocentre be $O(1, 2)$.$1$. The slope of $BC$ is $m_{BC} = \frac{5 - (-3)}{-2 - 4} = \frac{8}{-6} = -\frac{4}{3}$.Since the altitude $AO$ is perpendicular to $BC$,the slope of $AO$ is $m_{AO} = -\frac{1}{m_{BC}} = \frac{3}{4}$.The equation of line $AO$ passing through $A(h, k)$ and $O(1, 2)$ is $\frac{k - 2}{h - 1} = \frac{3}{4}$,which simplifies to $4(k - 2) = 3(h - 1) \Rightarrow 3h - 4k + 5 = 0$ ... $(i)$.$2$. The slope of $AC$ is $m_{AC} = \frac{k - 5}{h - (-2)} = \frac{k - 5}{h + 2}$.Since the altitude $BO$ is perpendicular to $AC$,the slope of $BO$ is $m_{BO} = \frac{2 - (-3)}{1 - 4} = \frac{5}{-3} = -\frac{5}{3}$.Therefore,$m_{AC} = -\frac{1}{m_{BO}} = \frac{3}{5}$.Equating the slopes: $\frac{k - 5}{h + 2} = \frac{3}{5}$ $\Rightarrow 5(k - 5) = 3(h + 2)$ $\Rightarrow 3h - 5k + 31 = 0$ ... $(ii)$.$3$. Solving equations $(i)$ and $(ii)$:Subtracting $(ii)$ from $(i)$: $(3h - 4k + 5) - (3h - 5k + 31) = 0$ $\Rightarrow k - 26 = 0$ $\Rightarrow k = 26$.Substituting $k = 26$ into $(i)$: $3h - 4(26) + 5 = 0$ $\Rightarrow 3h - 104 + 5 = 0$ $\Rightarrow 3h = 99$ $\Rightarrow h = 33$.Thus,the third vertex is $(33, 26)$.

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

Similar Questions

If $(a, b - c)$,$(b, c - a)$,and $(c, a - b)$ are the vertices of a triangle,then where does the centroid of the triangle lie?

If $P(2, 2)$,$Q(-2, 4)$,and $R(3, 4)$ are the vertices of $\triangle PQR$,then the equation of the median through vertex $R$ is $........$

If in triangle $ABC$,$A \equiv (1, 10)$,circumcentre $\equiv \left( -\frac{1}{3}, \frac{2}{3} \right)$ and orthocentre $\equiv \left( \frac{11}{3}, \frac{4}{3} \right)$,then the coordinates of the mid-point of the side opposite to $A$ are:

In $\Delta ABC$,the coordinates of $B$ are $(0, 0)$,$AB = 2$,$\angle ABC = \frac{\pi}{3}$,and the coordinates of the midpoint of $BC$ are $(2, 0)$. Find the centroid of the triangle.

If the vertices $P, Q, R$ of a triangle $PQR$ are rational points (points with rational coordinates),which of the following points is/are not necessarily a rational point?

Vedclass Test Series

Exam Paper Generator

Online Exam Module