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, …

Find the equation of line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0. Read More »

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 …

Find the coordinates of the point of intersecton of the lines 2x – y + 3 = 0 and x + y – 5 = 0. Read More »

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 = …

Find the equation of line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0. Read More »

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 …

If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\) Read More »

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 \(x\over a\) + \(y\over b\) = …

Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1) Read More »

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 …

If the line 2x + y = k passes through the point which divides the line segment joining the points (1,1) and (2,4) in the ratio 3:2, then k is equal to Read More »

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 + …

The x-coordinate of the incenter of the triangle that has the coordinates of mid-point of its sides as (0,1), (1,1) and (1,0) is Read More »

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 …

Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then Read More »

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 = …

If PS is the median of the triangle, with vertices of P(2,2), Q(6,-1) and R(7,3), then equation of the line passing through (1,-1) and parallel to PS is Read More »

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 = …

If \(\lambda x^2 – 10xy + 12y^2 + 5x – 16y – 3\) = 0 represents a pair of straight lines, then \(\lambda\) is equal to Read More »