The equations of two altitudes of an equilate | vedclass

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

Question

(B) Let the two given altitudes be $L_1: \sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $L_2: \sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The intersection of these t

Vedclass · Accepted Answer

$y = 10$

Vedclass · Answer

(B) Let the two given altitudes be $L_1: \sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $L_2: \sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The intersection of these two altitudes is the orthocenter of the triangle. Adding $L_1$ and $L_2$: $(\sqrt{3}x - y + 8 - 4\sqrt{3}) + (\sqrt{3}x + y - 12 - 4\sqrt{3}) = 0$. $2\sqrt{3}x - 4 - 8\sqrt{3} = 0$ $\Rightarrow 2\sqrt{3}x = 4 + 8\sqrt{3}$ $\Rightarrow x = \frac{2 + 4\sqrt{3}}{\sqrt{3}} = 4 + \frac{2}{\sqrt{3}}$. Subtracting $L_1$ from $L_2$: $(\sqrt{3}x + y - 12 - 4\sqrt{3}) - (\sqrt{3}x - y + 8 - 4\sqrt{3}) = 0$. $2y - 20 = 0 \Rightarrow y = 10$. The orthocenter is $(4 + \frac{2}{\sqrt{3}}, 10)$. In an equilateral triangle,the orthocenter is the same as the centroid. Since the third altitude must pass through the orthocenter $(x_0, y_0) = (4 + \frac{2}{\sqrt{3}}, 10)$,we check the given options. Substituting $(4 + \frac{2}{\sqrt{3}}, 10)$ into $y = 10$,we see it satisfies the equation. Thus,the equation of the third altitude is $y = 10$.

The equations of two altitudes of an equilateral triangle are $\sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $\sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The equation of the third altitude is

Similar Questions

The points $(1,3)$ and $(5,1)$ are opposite vertices of a diagonal of a rectangle. If the other two vertices lie on the line $y=2x+c$,then one of the vertices on the other diagonal is

Let $ABCD$ be a convex quadrilateral in which $AC = BD$,$AB = CD$,$\angle BAC = 70^{\circ}$ and $\angle BCD = 60^{\circ}$. The acute angle between $AC$ and $BD$ is (in $^{\circ}$)

If $ABCD$ is a quadrilateral,and the midpoints of consecutive sides $AB, BC, CD$,and $DA$ are joined by straight lines to form a quadrilateral $PQRS$,then $PQRS$ is always:

If the mid-points of the sides $BC$,$CA$ and $AB$ of a triangle $ABC$ are respectively $(2,1)$,$(-1,-2)$ and $(3,3)$,then the equation of the side $BC$ is

Let $A(h, k)$,$B(1, 1)$,and $C(2, 1)$ be the vertices of a right-angled triangle with $AC$ as its hypotenuse. If the area of the triangle is $1$ square unit,then the set of values which $k$ can take is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module