Which one of the following statements is true | vedclass

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

Question

(C) For option $D$,the homogeneous part of the given equation $xy - 3y^2 + y - 2x + 10 = 0$ is $xy - 3y^2 = 0$.To find the pair of lines through the o

Vedclass · Accepted Answer

The midpoints of the sides of a triangle are $(1, 2)$,$(3, 1)$,and $(5, 5)$. The orthocentre of the triangle has the coordinates $(3, 1)$.

Vedclass · Answer

(C) For option $D$,the homogeneous part of the given equation $xy - 3y^2 + y - 2x + 10 = 0$ is $xy - 3y^2 = 0$.To find the pair of lines through the origin perpendicular to these,we replace $x$ with $y$ and $y$ with $-x$ in the homogeneous part.Substituting these,we get $y(-x) - 3(-x)^2 = 0$,which simplifies to $-xy - 3x^2 = 0$,or $3x^2 + xy = 0$.Since the given option $D$ states $3y^2 + xy = 0$,it is incorrect.Evaluating option $A$: The lines $2x + 3y + 19 = 0$ and $9x + 6y - 17 = 0$ do not form a concyclic set with the axes.Evaluating option $B$: For a general triangle,the circumcentre,orthocentre,and centroid are collinear (Euler line),but the incentre is only collinear with them if the triangle is isosceles. The triangle with vertices $(1, 2), (4, 6), (-2, -1)$ is not isosceles.Evaluating option $C$: The vertices of the triangle are found by doubling the midpoints. Let midpoints be $M_1(1, 2), M_2(3, 1), M_3(5, 5)$. The vertices are $A(1+3-5, 2+1-5) = (-1, -2)$,$B(1+5-3, 2+5-1) = (3, 6)$,and $C(3+5-1, 1+5-2) = (7, 4)$. Calculating the orthocentre for these vertices confirms the statement is true.

Which one of the following statements is true?

Similar Questions

In $\triangle ABC$,let $AD, BE$ and $CF$ be the internal angle bisectors with $D, E$ and $F$ on the sides $BC, CA$ and $AB$ respectively. Suppose $AD, BE$ and $CF$ concur at $I$ and $B, D, I, F$ are concyclic,then $\angle IFD$ has measure $......$

If the straight lines $2x + 3y - 1 = 0$,$x + 2y - 1 = 0$,and $ax + by - 1 = 0$ form a triangle with the origin as the orthocentre,then $(a, b)$ is equal to

$A$ vertex of a square is $(3, 4)$ and one diagonal lies on the line $x + 2y = 1$. Find the equation of the second diagonal which passes through the given vertex.

The area of the parallelogram formed by the lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + d_1 = 0$,$a_2x + b_2y + c_2 = 0$,and $a_2x + b_2y + d_2 = 0$ is:

$A$ line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( \frac{1}{2}, 0 \right)$ and is perpendicular to $L$. Then the area of the triangle formed by the lines $L, L'$ and the $y$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module