L equiv 7x - y + 8 = 0 is one of the diagonal | vedclass

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

Question

(A) Let the vertices of the square be $A(-4, 5)$ and $B(3, 4)$. The diagonal $L$ is $7x - y + 8 = 0$. Since $A$ and $B$ are vertices of the square,the

Vedclass · Accepted Answer

$(0, 8), (-1, 1)$

Vedclass · Answer

(A) Let the vertices of the square be $A(-4, 5)$ and $B(3, 4)$. The diagonal $L$ is $7x - y + 8 = 0$. Since $A$ and $B$ are vertices of the square,the midpoint of the diagonal $AB$ is $M = (\frac{-4+3}{2}, \frac{5+4}{2}) = (-0.5, 4.5)$. Check if $M$ lies on $L$: $7(-0.5) - 4.5 + 8 = -3.5 - 4.5 + 8 = 0$. Thus,$AB$ is the other diagonal. The slope of $AB$ is $m_1 = \frac{4-5}{3-(-4)} = \frac{-1}{7}$. The slope of diagonal $L$ is $m_2 = 7$. Since $m_1 	imes m_2 = -1$,the diagonals are perpendicular,which is consistent with a square. The vertices on $L$ are equidistant from $M$ and at a distance equal to half the length of diagonal $AB$. Length $AB = \sqrt{(3 - (-4))^2 + (4 - 5)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{50} = 5\sqrt{2}$. Half-length $d = \frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}$. The line $L$ has direction vector $(1, 7)$. The unit vector is $u = \pm(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}}) = \pm(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}})$. The vertices on $L$ are $M \pm d \cdot u = (-0.5, 4.5) \pm \frac{5}{\sqrt{2}} \cdot (\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}) = (-0.5, 4.5) \pm (0.5, 3.5)$. Vertex $1: (-0.5 + 0.5, 4.5 + 3.5) = (0, 8)$. Vertex $2: (-0.5 - 0.5, 4.5 - 3.5) = (-1, 1)$. Thus,the vertices are $(0, 8)$ and $(-1, 1)$.

$L \equiv 7x - y + 8 = 0$ is one of the diagonals of a square for which $(-4, 5)$ and $(3, 4)$ are two vertices. Find the coordinates of the two vertices lying on the diagonal $L = 0$.

Similar Questions

The triangle formed by joining the points $P(2, 7)$,$Q(4, -1)$,and $R(-2, 6)$ is:

The two vertices of a triangle are $(2, -1)$ and $(3, 2)$,and the third vertex lies on the line $x + y = 5$. If the area of the triangle is $4$ square units,then the third vertex is:

$\Delta ABC$ has integral side lengths. The internal angle bisector of angle $B$ meets side $AC$ at $D$. If $AD = 3$ and $DC = 8$,then the perimeter of $\Delta ABC$ is:

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,then its locus is

Vedclass Test Series

Exam Paper Generator

Online Exam Module