Straight Line Questions

Solution : On solving the equations 4x + y – 1 = 0 and 7x – 3y – 35 = 0 by using point of intersection formula, we get x = 2 and y = -7 So, given lines intersect at (2, -7) Now, the equation of line joining the point (3, 5) and (2, […]

Solution : Solving simultaneously the equations 2x – y + 3 = 0 and x + y – 5 = 0, we obtain $x\over {5 – 3}$ = $y\over {3 + 10}$ = $1\over {2 + 1}$ $\implies$ $x\over 2$ = $y\over 13$ = $1\over 3$ $\implies$ x = $2\over 3$ , y = \(13\over

Solution : On solving the equations x – 7y + 5 = 0 and 3x + y = 0 by using point of intersection formula, we get x = $-5\over 22$ and y = $15\over 22$ So, given lines intersect at $({-5\over 22}., {15\over 22})$ Let the equation of the required line be x =

Solution : The given line is bx + ay – ab = 0 ………….(i) It is given that p = Length of the perpendicular from the origin to line (i) $\implies$ p = $|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}$ = $ab\over \sqrt{a^2+b^2}$ $\implies$ $p^2$ = $a^2b^2\over a^2+b^2$ $\implies$ $1\over p^2$ = $a^2+b^2\over a^2b^2$ $\implies$ \(1\over

Solution : We have line 12x – 5y + 9 = 0 and the point (2,1) Required distance = |$12*2 – 5*1 + 9\over {\sqrt{12^2 + (-5)^2}}$| = $|24-5+9|\over 13$ = $28\over 13$ Similar Questions If p is the length of the perpendicular from the origin to the line $x\over a$ + $y\over b$ =

Solution : Given line L : 2x + y = k passes through point (Say P) which divides the line segment (let AB) in ration 3:2, where A(1, 1) and B(2, 4). Using section formula, the coordinates of the point P which divides AB internally in the ratio 3:2 are P(\(3\times 2 + 2\times 1\over

Solution : Given mid-points of a triangle are (0,1), (1,1) and (1,0). So, by distance formula sides of the triangle are 2, 2 and $2\sqrt{2}$. x-coordinate of the incenter = $2*0 + 2\sqrt{2}*0 + 2*2\over {2 + 2 + 2\sqrt{2}}$ = $2\over {2+\sqrt{2}}$ Similar Questions Find the distance between the line 12x – 5y +

Solution : Since, ($\alpha$, -$\alpha$) lie on 4ax+2ay+c=0 and 5bx+2by+d=0. $\therefore$ 4a$\alpha$ + 2a$\alpha$ + c = 0 $\implies$ $\alpha$ = $-c\over 2a$  …..(i) Also, 5b$\alpha$ – 2b$\alpha$ + d = 0 $\implies$ $\alpha$ = $-d\over 3b$    …..(i) from equation (i) and (ii), $-c\over 2a$ = $-d\over 3b$ 3bc = 2ad Similar Questions Find

Solution : Since PS is the median, so S is the mid point of triangle PQR. So, Coordinates of S = (${7+6\over 2}, {3 – 1\over 2}$) = ($13\over 2$, 1) Slope of line PS = (1 – 2)/(13/2 – 2) = $-2\over 9$ Required equation passes through (1, -1) is y + 1 =

Solution : Comparing with $ax^2+2hxy+by^2+2gx+2fy+c$ = 0 Here a = $\lambda$, b = 12, c = -3, f = -8, g = 5/2, h = -5 Using condition $abc+2fgh-af^2-bg^2-ch^2$ = 0, we have $\lambda$(12)(-3) + 2(-8)(5/2)(-5) – $\lambda$(64) – 12(25/4) + 3(25) = 0 $\implies$  -36$\lambda$ + 200 – 64$\lambda$ – 75 + 75 =