A(1, -2), B(-2, 3), C(-1, -3) are the vertice | vedclass

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

Question

(A) The slope of $BC$ is $\frac{-3 - 3}{-1 - (-2)} = \frac{-6}{1} = -6$.The slope of the altitude $L_1$ from $A$ to $BC$ is the negative reciprocal of

Vedclass · Accepted Answer

$13l$

Vedclass · Answer

(A) The slope of $BC$ is $\frac{-3 - 3}{-1 - (-2)} = \frac{-6}{1} = -6$.The slope of the altitude $L_1$ from $A$ to $BC$ is the negative reciprocal of the slope of $BC$, which is $\frac{1}{6}$.The equation of $L_1$ passing through $A(1, -2)$ is $y - (-2) = \frac{1}{6}(x - 1) \Rightarrow y + 2 = \frac{x}{6} - \frac{1}{6} \Rightarrow y = \frac{x}{6} - \frac{13}{6}$ ... $(i)$The midpoint of $AB$ is $\left(\frac{1 - 2}{2}, \frac{-2 + 3}{2}\right) = \left(-\frac{1}{2}, \frac{1}{2}\right)$.The slope of $AB$ is $\frac{3 - (-2)}{-2 - 1} = \frac{5}{-3} = -\frac{5}{3}$.The slope of the perpendicular bisector $L_2$ is $\frac{3}{5}$.The equation of $L_2$ is $y - \frac{1}{2} = \frac{3}{5}(x + \frac{1}{2}) \Rightarrow y = \frac{3x}{5} + \frac{3}{10} + \frac{5}{10} \Rightarrow y = \frac{3x}{5} + \frac{4}{5}$ ... $(ii)$Equating $(i)$ and $(ii)$ to find the intersection point $(l, m)$:$\frac{l}{6} - \frac{13}{6} = \frac{3l}{5} + \frac{4}{5} \Rightarrow \frac{l}{6} - \frac{3l}{5} = \frac{4}{5} + \frac{13}{6} \Rightarrow \frac{5l - 18l}{30} = \frac{24 + 65}{30} \Rightarrow -13l = 89 \Rightarrow l = -\frac{89}{13}$.Substituting $l$ into $(i)$: $m = \frac{-89/13}{6} - \frac{13}{6} = \frac{-89}{78} - \frac{169}{78} = \frac{-258}{78} = -\frac{43}{13}$.Then $26m - 3 = 26(-\frac{43}{13}) - 3 = 2(-43) - 3 = -86 - 3 = -89$.Since $13l = -89$, we have $26m - 3 = 13l$.

$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the altitude drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3$ is equal to:

Similar Questions

If the orthocenter of a triangle is $(1, 1)$ and the circumcenter is $\left( \frac{3}{2}, \frac{3}{4} \right)$,find its centroid.

Let $ABC$ be an equilateral triangle with orthocenter at the origin and the side $BC$ on the line $x+2\sqrt{2}y=4$. If the coordinates of the vertex $A$ are $(\alpha, \beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta|$ is

If the midpoints of the sides of a triangle are $(-2, 3), (4, -3)$ and $(4, 5)$,then the centroid of the triangle is

The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is

Two vertices of a triangle are $(5, -1)$ and $(-2, 3)$. If the orthocentre is the origin,then the coordinates of the third vertex are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module