Question 17 of 25advanced🔍 AnalyzeLong Answer4 marks

Are the points M(−3,−4)M (-3, -4), A(0,0)A (0, 0) and G(6,8)G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.

Correct Answer

Yes, the points M(−3,−4)M (-3, -4), A(0,0)A (0, 0), and G(6,8)G (6, 8) lie on the same straight line.

A method to check if three points are collinear without plotting is by calculating the distances between them. If the sum of the distances of two smaller segments equals the distance of the largest segment, the points lie on the same straight line. Alternatively, one can check if the slopes of the line segments joining the points are equal.

Step 1: Calculate distance MA=[(0−(−3))2+(0−(−4))2]=(32+42)=(9+16)=5MA = \sqrt{[(0 - (-3))^2 + (0 - (-4))^2]} = \sqrt{(3^2 + 4^2)} = \sqrt{(9 + 16)} = 5 units.

Step 2: Calculate distance AG=[(6−0)2+(8−0)2]=(62+82)=(36+64)=10AG = \sqrt{[(6 - 0)^2 + (8 - 0)^2]} = \sqrt{(6^2 + 8^2)} = \sqrt{(36 + 64)} = 10 units.

Step 3: Calculate distance MG=[(6−(−3))2+(8−(−4))2]=(92+122)=(81+144)=15MG = \sqrt{[(6 - (-3))^2 + (8 - (-4))^2]} = \sqrt{(9^2 + 12^2)} = \sqrt{(81 + 144)} = 15 units.

Step 4: Since MA+AG=5+10=15MA + AG = 5 + 10 = 15 units, and MG=15MG = 15 units, we have MA+AG=MGMA + AG = MG. Thus, the points are collinear.

Exercise: END-OF-CHAPTER EXERCISES | Q: *6 | (Page 12)
For More Understanding

Explanation

The question requires verifying the collinearity of three points without plotting. The distance method is highly appropriate here, extending the concept of finding lengths (like the radius of a circle from the origin or sides of a triangle) mentioned in the context. By calculating the distances MAMA, AGAG, and MGMG, we can verify if the sum of the two shorter distances equals the longest distance. The slope method (comparing slopes of MAMA and AGAG) is another valid approach, as equal slopes passing through a common point indicate a single straight line.

Solution Steps

  1. Step 1: Calculate the distance between M(−3,−4)M(-3, -4) and A(0,0)A(0, 0).

    MA=[(0−(−3))2+(0−(−4))2]=(32+42)=25=5 unitsMA = \sqrt{[(0 - (-3))^2 + (0 - (-4))^2]} = \sqrt{(3^2 + 4^2)} = \sqrt{25} = 5 \text{ units}

  2. Step 2: Calculate the distance between A(0,0)A(0, 0) and G(6,8)G(6, 8).

    AG=[(6−0)2+(8−0)2]=(62+82)=100=10 unitsAG = \sqrt{[(6 - 0)^2 + (8 - 0)^2]} = \sqrt{(6^2 + 8^2)} = \sqrt{100} = 10 \text{ units}

  3. Step 3: Calculate the distance between M(−3,−4)M(-3, -4) and G(6,8)G(6, 8).

    MG=[(6−(−3))2+(8−(−4))2]=(92+122)=225=15 unitsMG = \sqrt{[(6 - (-3))^2 + (8 - (-4))^2]} = \sqrt{(9^2 + 12^2)} = \sqrt{225} = 15 \text{ units}

  4. Step 4: Check the condition for collinearity.

    MA+AG=5+10=15 unitsMA + AG = 5 + 10 = 15 \text{ units}

    Since MA+AG=MGMA + AG = MG (15=1515 = 15), the points MM, AA, and GG lie on the same straight line.

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