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 : Given the equation of ellipse is \(x^2\over 16\) + \(y^2\over 9\) = 1 Here, a = 4, b = 3, e = \(\sqrt{1-{9\over 16}}\) = \(\sqrt{7\over 4}\) \(\therefore\) foci is (\(\pm ae\), 0) = (\(\pm\sqrt{7}\), 0) \(\therefore\) Radius of the circle, r = \(\sqrt{(ae)^2+b^2}\) r = \(\sqrt{7+9}\) = \(\sqrt{16}\) = 4 Now, equation …
Solution : Let the equation of circle be \((x-3)^2 + (y-0)^2 + \lambda y\) = 0 As it passes through (1, -2) \(\therefore\) \((1-3)^2 + (-2)^2 + \lambda (-2)\) = 0 \(\implies\) 4 + 4 – 2\(\lambda\) = 0 \(\implies\) \(\lambda\) = 4 \(\therefore\) Equation of circle is \((x-3)^2 + y^2 + 4y\) = 0 …
Solution : for parabola, \(y^2\) = 4x Let y = mx + \(1\over m\) is tangent line and it touches the parabola \(x^2\) = -32. \(\therefore\) \(x^2\) = -32(mx + \(1\over m\)) \(\implies\) \(x^2 + 32mx + {32\over m}\) = 0 Now, D = 0 because it touches the curve. \(\therefore\) \((32m)^2 – 4.{32\over m}\) …
The slope of the line touching both the parabolas \(y^2\) = 4x and \(x^2\) = -32 is Read More »
Solution : Let coordinates of the center of T be (0, k). Distance between their center is k + 1 = \(\sqrt{1 + (k – 1)^2}\) where k is radius of circle T and 1 is radius of circle C, so sum of these is distance between their centers. \(\implies\) k + 1 = \(\sqrt{k^2 + …
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 : We have [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = {(\(\vec{a}\) + \(\vec{b}\))\(\times\)(\(\vec{b}\) + \(\vec{c}\))}.(\(\vec{c}\) + \(\vec{a}\)) = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{b}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) {\(\vec{b}\)\(\times\)\(\vec{b}\) = 0} = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) = (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{b}\)).\(\vec{a}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{a}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{a}\) = …
Solution : \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) \(\vec{c}\perp\vec{a}\) , \(\vec{c}\perp\vec{b}\) and \(\vec{a}\perp\vec{b}\), \(\vec{a}\perp\vec{c}\) \(\implies\) \(\vec{a}\perp\vec{b}\), \(\vec{b}\perp\vec{c}\) and \(\vec{c}\perp\vec{a}\) \(\implies\) \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors. Again, \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) |\(\vec{a}\times\vec{b}\)| = |\(\vec{c}\)| and |\(\vec{b}\times\vec{c}\)| = |\(\vec{a}\)| \(\implies\) \(|\vec{a}||\vec{b}|sin{\pi\over 2}\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|sin{\pi\over 2}\) = |\(\vec{a}\)| …
Solution : Unit vectors perpendicular to \(\vec{a}\) & \(\vec{b}\) = \(\pm\)\(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) \(\therefore\) \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}\) = \(-5\hat{i} – 5\hat{j} – 5\hat{k}\) \(\therefore\) Unit Vectors = \(\pm\) \(-5\hat{i} – 5\hat{j} – 5\hat{k}\over 5\sqrt{3}\) Hence the required …
