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Dev Puri
Subject: Maths
, asked on 30/3/18
Is there a chapter of mensuration in class 11 or 12
Answer
1
Affan
Subject: Maths
, asked on 20/3/18
Question 22 ans please
Q22. If each of the points (x
_{1}
, 4) , (–2, y
_{1}
) lies on the joining the points (2, –1). ( 5, –3), then the point P (x
_{1}
, y
_{1}
) lies on the line
(a) 6 (x + y) – 25 = 0 (b) 2x + 6y + 1 = 0
(c) 2x + 3 y – 6 = 0 (d) 6 ( x + y ) + 25 = 0
Answer
1
Ipsita Dash
Subject: Maths
, asked on 20/3/18
Write the area of the triangle formed by the coordinate axes and the line (secA-tanA)x + (secA+tanA)y =2.
Answer
1
Priyansh
Subject: Maths
, asked on 19/3/18
Q. 2x+y=7 is one of the bisectors of the angles between the lines represented by
$\alpha {\left(x-2\right)}^{2}+2\gamma \left(x-2\right)\left(y-3\right)+\beta {\left(y-3\right)}^{2}=0$
. Then find the intercepts made by the other bisector on x-axis and y-axis respectively.
Answer
1
Priyansh
Subject: Maths
, asked on 19/3/18
Q6 please answer fast
Answer
1
Yagyam Aggarwal
Subject: Maths
, asked on 18/3/18
Pls answer 83 question:
83. The equation of lines joining the origin and points of intersection of x
^{2}
+y
^{2}
-2xy=4 and 3x
^{2}
+5y
^{2}
-xy=7 is
(1) x
^{2}
+y
^{2}
-xy=0
(2) 5x
^{2}
+2y
^{2}
-13xy=0
(3) 5x
^{2}
+y
^{2}
-6xy=0
(4) 5x
^{2}
+13y
^{2}
-18xy=0
Answer
2
Chharishma.r.nayaka
Subject: Maths
, asked on 18/3/18
Plz answer fast...
Q.35. Find the centre and radius of the circle
${x}^{2}+{y}^{2}+8x+10y-8=0$
Answer
1
Erw Trjerth
Subject: Maths
, asked on 16/3/18
Solve this:
Answer
1
Yagyam Aggarwal
Subject: Maths
, asked on 13/3/18
If
x
= 3 + 4 cos θ,
y
= 2 + 3 sin θ then locus of
P
(
x, y
) is
(1) Straight line (2) Circle (3) Parabola (4) Ellipse
Answer
1
Yagyam Aggarwal
Subject: Maths
, asked on 13/3/18
Please answer 87th question
$\mathbf{87}\mathbf{.}\mathbf{}\mathbf{}Thelocusofthemidpointsoftheportionofthe\mathrm{tan}gentstotheellipse\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1interceptedbetweentheaxesis\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(1\right)\frac{16}{{x}^{2}}+\frac{9}{{y}^{2}}=4\left(2\right)\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right)\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=16\left(4\right)\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=25$
Answer
1
Drishti
Subject: Maths
, asked on 6/3/18
Solve 14
Answer
1
Drishti
Subject: Maths
, asked on 6/3/18
Solve 3 fast
Answer
1
Venkatesan Rs
Subject: Maths
, asked on 6/3/18
A traffic policeman at P(3,5) wants to reach the nearest point Q on the road along the line 5x+6y=106 as early as possible.
1.Locate the position of Q
2.Find the distance PQ
Answer
1
Drishti
Subject: Maths
, asked on 5/3/18
Q. The hypotenuse side of isosceles right triangle lies along the line 2x-y=4 and vertex opposite to hypotenuse is (1,5). Obtain the equations of other two sides.
Answer
1
Drishti
Subject: Maths
, asked on 5/3/18
Solve the ques6
Answer
1
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What are you looking for?

Q22. If each of the points (x

_{1}, 4) , (–2, y_{1}) lies on the joining the points (2, –1). ( 5, –3), then the point P (x_{1}, y_{1}) lies on the line(a) 6 (x + y) – 25 = 0 (b) 2x + 6y + 1 = 0

(c) 2x + 3 y – 6 = 0 (d) 6 ( x + y ) + 25 = 0

$\alpha {\left(x-2\right)}^{2}+2\gamma \left(x-2\right)\left(y-3\right)+\beta {\left(y-3\right)}^{2}=0$ . Then find the intercepts made by the other bisector on x-axis and y-axis respectively.

83. The equation of lines joining the origin and points of intersection of x

^{2}+y^{2}-2xy=4 and 3x^{2}+5y^{2}-xy=7 is(1) x

^{2}+y^{2}-xy=0(2) 5x

^{2}+2y^{2}-13xy=0(3) 5x

^{2}+y^{2}-6xy=0(4) 5x

^{2}+13y^{2}-18xy=0Q.35. Find the centre and radius of the circle ${x}^{2}+{y}^{2}+8x+10y-8=0$

x= 3 + 4 cos θ,y= 2 + 3 sin θ then locus ofP(x, y) is(1) Straight line (2) Circle (3) Parabola (4) Ellipse

$\mathbf{87}\mathbf{.}\mathbf{}\mathbf{}Thelocusofthemidpointsoftheportionofthe\mathrm{tan}gentstotheellipse\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1interceptedbetweentheaxesis\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(1\right)\frac{16}{{x}^{2}}+\frac{9}{{y}^{2}}=4\left(2\right)\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(3\right)\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=16\left(4\right)\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=25$

1.Locate the position of Q

2.Find the distance PQ

Q. The hypotenuse side of isosceles right triangle lies along the line 2x-y=4 and vertex opposite to hypotenuse is (1,5). Obtain the equations of other two sides.