26. AB is a line which cuts off positive intercepts OA and OB on the respective axes of reference. If
P(4, 2) be a point on the line segment AB such that PA: PB = 4:5 then find the equation and the
slope of the line AB.
27. Let P(5,0), Q(-2, 2) and R(3, 4) be the vertices of the APQR. Find the equation of the median
through (a) P, (b) Q, and (c) R.
28. Let ABC be a triangle whose centroid is G. If B =(-2, 0), C =(3,1) and G =(1,5) then find the
equations of the sides (a) AB, (b) AC.
29. (a) Let the lines AB and AC have the equations 3x – 4y +1 = 0 and x + y = 2 respectively. Find
the equation of the line AP, where P has the coordinates (-2,5).
(b) Find the point midway between the point (-1, 3) and the point of intersection of the lines
4x + y =10 and 2x + 3y = 8.
30. (a) The sides of a rectangle have the equations x + 2 = 0, y + 3 = 0, x = 4 and y=1. Find the
equations of its diagonals.
(b) Prove that the lines x = 2, x +1 = 0, y = 0 and y = 3 form a square. Find the equations of its
diagonals.
Answers
Answer:
Find the equation of a straight line passing through a point (3,-2) and cuts off positive intercepts on the x-axis and y-axis which are in ratio 4:3.
Step-by-step explanation:
The equation of the required straight line :
• In general form is ; 3x + 4y - 1 = 0 .
• In x,y-intercept form is ; x/(1/3) + y/(1/4) = 1 .
Note:
• If any straight line (or any other curve) passes through a given point, then the coordinates of the given point will satisfy the equation of that straight line (or that curve).
• The equation of a straight line in general form is given by ; Ax + By + C = 0.
• The equation of a straight line in slope, y-intercept form is given by ; y = mx + c , where , m is the slope and c is the y-intercept of the straight line.
• The equation of a straight line in x, y-intercept form is given by; x/a + b/y = 1 , where a is the x-intercept and b is the y-intercept.
Also, the point of interception on x-axis is (a,0) and the point of interception on y-axis is (0,b).
Solution:
It is given that;
The required straight line cuts off positive intercepts on the x-axis and y-axis which are in ratio 4:3. ie; a:b = 4:3.
Thus;
Let a = 4p and b = 3p.
Therefore,
The equation of the required straight line will be given by;
=> x/a + y/b = 1
=> x/4p + y/3p = 1
=> (3x + 4y)/12p = 1
=> 3x + 4y = 12p ----------(1)
Also,
It is given that, the required straight line passes through the given point (3,-2).
Thus;
The coordinates of the given point (ie; x = 3 and
y = -2) will satisfy the equation of the required straight line (ie; eq-1).
Now,
Putting x = 3 and y = -2 in eq-1 , we get;
=> 3x + 4y = 12p
=> 3•3 + 4•(-2) = 12p
=> 9 - 8 = 12p
=> 1 = 12p
=> p = 1/12
Now,
Putting p = 1/12 in eq-1 , we get;
=> 3x + 4y = 12p
=> 3x + 4y = 12•(1/12)
=> 3x + 4y = 12/12
=> 3x + 4y = 1
=> 3x + 4y - 1 = 0
This is the equation of the required line in general form.
Also,
=> 3x + 4y = 1
=> x/(1/3) + y/(1/4) = 1