Question 1: Reduce the equation to:

(i) slope-intercept form and find slope and y-intercept (ii) intercept form and find intercept on the axes; (iii) the normal form and find and .

Answer:

i) Given

Therefore Slope and y-intercept

ii) Given

Therefore x-intercept and y-intercept

iii) Given

Therefore and

Since the coefficient of both and are negative, lies in the Quadrant.

Question 2: Reduce the following equations to the normal form and find and in each case:

i) ii) iii)

iv) v)

Answer:

i) Given

Therefore and

Since the coefficient of both and are positive, lies in the Quadrant.

ii) Given

Therefore and

Since the coefficient of both and are negative, lies in the Quadrant.

iii) Given

Therefore and

Since the coefficient of both is negative and y is positive, lies in the Quadrant.

iv) Given

Therefore and

v) Given

Therefore and

Question 3: Put the equation to the slope intercept form and find its slope and y-intercept.

Answer:

Given

Comparing with where slope and y-intercept

Therefore and y-intercept

Question 4: Reduce the lines and to the normal form and hence find which line is nearer to the origin.

Answer:

Line 1: Given

Therefore

Line 2: Given

Therefore

Since , we can say that is nearer to the origin as compared to

Question 5: Show that the origin is equidistant from the lines ; and .

Answer:

Line 1: Given

Therefore

Line 2: Given

Therefore

Line 3: Given

Therefore

Since , the given lines ; and are equidistance from the origin.

Question 6: Find the values of and , if the equation is the normal form of the line .

Answer:

Given

Therefore and

Question 7: Reduce the equation to the intercept form and find the and intercepts.

Answer:

Given equation

Hence, x-intercept and y-intercept

Question 8: The perpendicular distance of a line from the origin is units and its slope is . Find the equation of the line.

Answer:

Given

Therefore equation of line:

Comparing with where slope and y-intercept

Slope

Therefore the equation of line is