Chapter 14: Problem 1210
If the length of perpendicular drawn from \((5,0)\) on \(\mathrm{kx}+4 \mathrm{y}=20\) is 1, then \(\mathrm{k}=\ldots \ldots \ldots\) (a) \(3,(16 / 3)\) (b) \(3,-(16 / 3)\) (c) \(-3,(16 / 3)\) (d) \(-3,-(16 / 3)\)
Chapter 14: Problem 1210
If the length of perpendicular drawn from \((5,0)\) on \(\mathrm{kx}+4 \mathrm{y}=20\) is 1, then \(\mathrm{k}=\ldots \ldots \ldots\) (a) \(3,(16 / 3)\) (b) \(3,-(16 / 3)\) (c) \(-3,(16 / 3)\) (d) \(-3,-(16 / 3)\)
All the tools & learning materials you need for study success - in one app.
Get started for freeThe in centre of a triangle whose vertices \(\mathrm{A}(2,4), \mathrm{B}(2,6)\) and \(\mathrm{C}(2+\sqrt{3}, 5)\) is.... (a) \([2+(1 / \sqrt{3}), 5]\) (b) \([1+\\{1 /(2 \sqrt{3})\\},(5 / 2)]\) (c) \((2,5)\) (d) None of these
The equations of two straight lines which are parallel to \(\mathrm{x}+7 \mathrm{y}+2=0\) and at unit distance from the point \((1,-1)\) are (a) \(x+7 y+6 \pm 4 \sqrt{2}=0\) (b) \(x+7 y+6 \pm 5 \sqrt{2}=0\) (c) \(2 \mathrm{x}+7 \mathrm{y}+6 \pm 5 \sqrt{2}=0\) (d) \(x+y+6 \pm 5 \sqrt{2}=0\)
\(\mathrm{A}(4,0), \mathrm{B}(0,3), \mathrm{C}(6,1)\) be vertices of triangle \(\mathrm{ABC}\). Slope of bisector of angle \(\mathrm{C}\) will be (a) \(3 \sqrt{2}-7\) (b) \(5 \sqrt{2}-7\) (c) \(6 \sqrt{2}-7\) (d) none
The nearest point on the line \(\mathrm{x}-3 \mathrm{y}+25=0\) from the origin is \(\ldots \ldots\) (a) \((-4,5)\) (b) \((-4,3)\) (c) \((4,3)\) (d) None of these
In a triangle \(\mathrm{ABC}\), coordinates of \(\mathrm{A}\) are \((1,2)\) and the equations of the medians through \(\mathrm{B}\) and \(\mathrm{C}\) are \(\mathrm{x}+\mathrm{y}=5\) and \(\mathrm{x}=4\) respectively. Then coordinates of \(\mathrm{B}\) and \(\mathrm{C}\) will be (a) \((-2,7),(4,3)\) (b) \((7,-2),(4,3)\) (c) \((2,7),(-4,3)\) (d) \((2,-7),(3,-4)\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.