Straight Lines

Step 1: Find midpoints of opposite sides.
Step 2: Diagonals pass through midpoints.

Step 1: Standard form. 2x-5y+22=0
Step 2: Distance formula. d=|Ax₁+By₁+C|√(A²+B²)=5

Step 1: Equate coefficients. Set coefficients of a and b separately to zero.
Step 2: Solve simultaneous equations. Intersection point is obtained.

Given equations are
and


For acute angle bisector

Step 1: Use the formula for the distance from a point to a line.
The formula for the perpendicular distance from the origin to the line is given by: Substituting the given equation, the correct coordinates of the nearest point are .
Thus, the correct answer is (1).

Step 1: Analyze the equation.
The given equation represents a condition for the concurrent lines, and solving it gives the points of concurrency.
Step 2: Conclusion.
The points of concurrency are and . Final Answer:

Step 1: Analyze the line equation.
The line passing through has the general equation , which intersects the axes at and .
Step 2: Use the area formula.
The area of the triangle formed by the intercepts is given by: Thus, the quadratic equation becomes . Final Answer:

Step 1: Analyze the system of equations.
For the two lines to be concurrent, the system of equations must have a common solution. This happens when .
Step 2: Conclusion.
The lines are concurrent when . Final Answer:

We have = length of perpendicular from on

i.e.

= Length of the perpendicular from on


Now,

Step 1: Solve first two equations to find point of intersection.
Step 2: Substitute into third equation to get .

Step 1: Perpendicular distance from origin to is .
Step 2: Compute and using the given equations.
Step 3: Substitution gives:

Step 1: Concurrency implies the determinant of coefficients is zero.
Step 2: Hence satisfy a linear relation.
Step 3: Therefore the points are collinear.

Step 1: Check condition ab=h², so equation represents pair of parallel lines.
Step 2: Distance between them evaluates to (5)/(2).

Step 1: Reflect (5,3) in x-axis to (5,-3).
Step 2: Find intersection of line joining (1,2) and (5,-3) with x-axis.
Step 3: Equation gives x=(13)/(5).

We have , . Thus Hence the equation represents a pair of straight lines. Again the lines are parallel. The distance between them

Let the coordinates of point be Let be the normal. Let So, Slope of line ....(i) Slope of line .... (ii) From (i) and (ii), Hence, the image is

Radius in all values

Let be the point dividing the join of and in the ratio internally, then ..(i) (ii) Squaring and adding (i) and (ii), we get the required locus , which is a circle.

Step 1: A line of the system is (2x+y-3)+λ(3x+2y-5)=0.
Step 2: Distance from point (4,-3) is maximized when numerator is maximum.
Step 3: This occurs for the line 7x+y-8=0.

Slopes are -b/a and -a/b. tanθ=|(m₁-m₂)/(1+m₁m₂)| =(|b²-a²|)/(2ab).

If lines of the form ax+by=1 are concurrent, then the corresponding points (a,b) lie on a straight line. Hence the given points are collinear.

Answer (c) The lines intersect in unique point

Step 1: A general line of the system is (2a+3b)x+(a+2b)y-(3a+5b)=0 Step 2: Distance of a line Ax+By+C=0 from point (x₀,y₀) is: D=frac|Ax₀+By₀+C|√(A²+B²) Step 3: Evaluating distances of given options from (4,-3), the maximum distance is obtained for: 3x-4y+1=0 Hence, option (D) is correct.

Step 1: Identify the line equation form
The equation is: This is in slope-intercept form: where:
- : slope
- : y-intercept

Step 2: Extract the slope
Compare: The slope is 3.