NCERT Class 11-Math՚s: Exemplar Chapter – 10 Straight Lines Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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10.1. 8 Intersection of two given lines

Two lines and are

(i) Intersecting

(ii) Parallel and distinct if

(ii)

Remarks

(i) The points and are on the same side of the line or on the opposite side of the line , if and are of the same sign or of opposite signs respectively.

(ii) The condition that the lines and are perpendicular is .

(iii) The equation of any line through the point of intersection of two lines and is . The value of is determined from extra condition given in the problem.

10.2 Solved Examples

Short Answer Type

Question 1:

Find the equation of a line which passes through the point and makes an angle of with the positive direction of axis.

Answer:

Here the slope of the line is d the given point is . Therefore, using point slope formula of the equation of a line, we have

Question 2:

Find the equation of the line where length of the perpendicular segment from the origin to the line is and the inclination of the perpendicular segment with the positive direction of x-axis is .

Answer:

The normal form of the equation of the line is ere and . Therefore, the equation of the line is

Question 3:

Prove that every straight line has an equation of the form , where and are constants.

Answer:

Proof Given a straight line, either it cuts the y-axis, or is parallel to or coincident with it. We know that the equation of a line which cuts the y-axis (i.e.. , it has y-intercept) can be put in the form ; further, if the line is parallel to or coincident with the , its equation is of the form , where in the case of coincidence. Both of these equations are of the form given in the problem and hence the proof.

Question 4:

Find the equation of the straight line passing through and perpendicular to the line .

Answer:

Let be the slope of the line whose equation is to be found out which is perpendicular to the line . The slope of the given line is . Therefore, using the condition of perpendicularity of lines, we have or (Why?)

Hence, the required equation of the line is or .

Question 5:

Find the distance between the lines and .

Answer:

The equations of lines and may be rewritten as and

Since, the slope of these lines are same and hence they are parallel to each other. Therefore, the distance between them is given by

Question 6:

Show that the locus of the mid-point of the distance between the of the variable line is where p is a constant.

Answer:

Changing the given equation of the line into intercept form, we have

which gives the coordinates , where the line intersects x-axis and y-axis, respectively.

Let denote the mid-point of the line segment joining the points

Then

This gives

Squaring and adding we get

Therefore, the required locus is

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