If a diagonal of a square is along the line 8 | vedclass

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

Question

(B) The sides of a square make an angle of $45^{\circ}$ with the diagonal. The slope of the given diagonal $8x - 15y = 0$ is $m_1 = \frac{8}{15}$.Let

Vedclass · Accepted Answer

$23x - 7y - 9 = 0, 7x + 23y - 53 = 0$

Vedclass · Answer

(B) The sides of a square make an angle of $45^{\circ}$ with the diagonal. The slope of the given diagonal $8x - 15y = 0$ is $m_1 = \frac{8}{15}$.Let the slope of the sides passing through $(1, 2)$ be $m$. Using the formula $	an(45^{\circ}) = |\frac{m - m_1}{1 + m \cdot m_1}|$,we get:$1 = |\frac{m - 8/15}{1 + 8m/15}|$$1 = |\frac{15m - 8}{15 + 8m}|$This gives two cases:Case $1$: $15m - 8 = 15 + 8m$ $\Rightarrow 7m = 23$ $\Rightarrow m = \frac{23}{7}$.The equation of the side is $y - 2 = \frac{23}{7}(x - 1)$ $\Rightarrow 7y - 14 = 23x - 23$ $\Rightarrow 23x - 7y - 9 = 0$.Case $2$: $15m - 8 = -(15 + 8m)$ $\Rightarrow 15m - 8 = -15 - 8m$ $\Rightarrow 23m = -7$ $\Rightarrow m = -\frac{7}{23}$.The equation of the side is $y - 2 = -\frac{7}{23}(x - 1)$ $\Rightarrow 23y - 46 = -7x + 7$ $\Rightarrow 7x + 23y - 53 = 0$.

If a diagonal of a square is along the line $8x - 15y = 0$ and one of its vertices is $(1, 2)$,then the equations of the sides of the square passing through this vertex are

Similar Questions

If $(4,3)$ and $(1,-2)$ are the end points of a diagonal of a square,then the equation of one of its sides is

If the orthocenter of the triangle formed by the lines $2x + 3y - 1 = 0$,$x + 2y + 1 = 0$,and $ax + by - 1 = 0$ lies at the origin,then $\frac{1}{a} + \frac{1}{b} =$

The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

The circumcentre of the triangle formed by the lines $x+y+2=0, 2x+y+8=0$ and $x-y-2=0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module