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

Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a^2$ is equal to

Let the points $\left(\frac{11}{2}, \alpha\right)$ lie on or inside the triangle with sides $x + y = 11$,$x + 2y = 16$,and $2x + 3y = 29$. Then the product of the smallest and the largest values of $\alpha$ is equal to:

The set of values that $\beta$ can assume so that the point $(0, \beta)$ lies on or inside the triangle formed by the lines $3x+y+2=0$,$2x-3y+5=0$,and $x+4y-14=0$ is

The diagonal passing through the origin of a quadrilateral formed by $x = 0, y = 0, x + y = 1$ and $6x + y = 3$ is

If in a parallelogram $ABDC$,the coordinates of $A, B$ and $C$ are respectively $(1, 2), (3, 4)$ and $(2, 5)$,then the equation of the diagonal $AD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module