Question
Find the area of triangle formed by the points A(1, 2),
B(4, 6), and C(5, 2).Solution
Area = (1/2) | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) | = (1/2) | 1(6 - 2) + 4(2 - 2) + 5(2 - 6) | = (1/2) | 1(4) + 0 + 5(-4) | = (1/2) | 4 - 20 | = (1/2) × 16 = 8 sq. units.
The slope of the equation 24x+ 8y =56 is
The length of a line segment A, B is 10 units. The coordinates of point A are(2,3) and the value of ordinate of point B is 5. Then find the value of abs...
If (5√P - 7√Q) = 5, [1.5P = 4Q-(R/3)+9] and (√P/√Q) = 1.6, then find out the value of ‘R’.
If, x - y radic; 3 = 23 and radic; 3x + y = 35
Find the angle between both the lines when they intersect each other.
...I. p2 – 5p + 6 = 0
II. 36q2 = 81
...I. p² - 365 = 364
II. q - √ 529 = √ 169
Find the value of 'a' and 'b' which satisfy the following equations:
9a + 7b = 30
4a - 5b = 62
The ratio of roots of an equation
ax²+bx+b = 0 is p : q
Then find, √p/√q + √q/√p + √p/√q ?
If x - √3 y = 8 Find the slope in angle?
I. 8/(√x) + 6/(√x) = √x
II. y ³- (14)7/2 /(√y) = 0
