Question
The condition for the lines 
and 
to be perpendicular is
The correct answer is:
Related Questions to study
If f : R →R; f(x) = sin x + x, then the value of 
(f-1 (x)) dx, is equal to
Here we used the concept of integration and the inverse functions to solve the question. Finding an antiderivative of a function is the procedure known as integration. The process of adding the slices to complete it is comparable. The process of integration is the opposite of that of differentiation. So the final answer is 
.
If f : R →R; f(x) = sin x + x, then the value of (f-1 (x)) dx, is equal to
Here we used the concept of integration and the inverse functions to solve the question. Finding an antiderivative of a function is the procedure known as integration. The process of adding the slices to complete it is comparable. The process of integration is the opposite of that of differentiation. So the final answer is .
The polar equation of the straight line with intercepts 'a' and 'b' on the rays 
and 
respectively is
The polar equation of the straight line with intercepts 'a' and 'b' on the rays and respectively is
The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is
The polar equation of the straight line parallel to the initial line and at a distance of 4 units above the initial line is
The polar equation of 
axy is
The polar equation of axy is
If x, y, z are integers and x 
0, y 1, z 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -
If x, y, z are integers and x 0, y 1, z 2, x + y + z = 15, then the number of values of the ordered triplet (x, y, z) is -
If 
then the equation whose roots are 
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the equation is 
.
If then the equation whose roots are
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, the equation is .
Let p, q 
{1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, total 7 seven solutions are possible so 7 equations can be formed.
Let p, q {1, 2, 3, 4}. Then number of equation of the form px2 + qx + 1 = 0, having real roots, is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of discriminant and used it in the solution to find the intervals. Therefore, total 7 seven solutions are possible so 7 equations can be formed.
ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
Here we used the concept of quadratic equations and solved the problem. The concept of sum of roots and product of roots was also used here. Therefore, 0 < ∣α∣ < β.
ax2 + bx + c = 0 has real and distinct roots null. Further a > 0, b < 0 and c < 0, then –
Here we used the concept of quadratic equations and solved the problem. The concept of sum of roots and product of roots was also used here. Therefore, 0 < ∣α∣ < β.
The cartesian equation of 
is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is 
.
The cartesian equation of is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is .
The castesian equation of 
is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is 
.
The castesian equation of is
Here we used the concept of quadratic equations and solved the problem. We also understood the concept of trigonometric ratios and used the formula to find the equation. So the equation is .
is
